Prove that if a² + ab + b² = 0 then a = 0 and b = 0

[CynicusRex]

Homework Statement

Prove that if a² + ab + b² = 0 then a = 0 and b = 0
Hint: Recall the factorization of a³-b³. (Another solution will be discussed later when speaking about quadratic equations.)

Homework Equations

a² + ab + b² is close to a² + 2ab + b² = (a+b)²
a³-b³=(a-b)(a²+ab+b²)

The Attempt at a Solution

a² + ab + b² = 0
a² + ab + b² + ab = ab
a² + 2ab + b² = ab
(a+b)² = ab
(a+b)² - ab = 0

Here I figured, to get zero after a subtraction, both terms need to be equal. An addition of two numbers is the same to the multiplication of those two numbers only when a,b = 0 or a,b = 2. (0+0=0*0 and 2+2=2*2)
But since the first term is squared, (2+2)² - 2*2 = 16 - 4 ≠0. This leaves a,b = 0 as the only option to get a zero.

A second try was
a² + ab + b² = 0
a² + ab + b² + ab = ab
a² + 2ab + b² = ab
(a+b)² = ab
(a+b)² - ab = 0
(b+b)² - bb = 0 (given that a=b, and same works for b=a)
(2b)² - b² = 0
(2b-b)(2b+b) = 0
b*b = 0
b = 0

A third unsuccessful attempt
a² + ab + b² = 0
a² + ab + b² + ab = ab
a² + 2ab + b² = ab
(a+b)² = ab
a+b = √(ab)
a+b = √a√b
(a+b) - (√a√b) = 0

The solution:
a³-b³=(a-b)(a²+ab+b²), if (a²+ab+b²)=0, then a³-b³=0 or
a³ = b³ which can only happen when a=b. If a=b, then 0=a²+ab+b²=3b² which implies that
b=0 and a=b=0.

I have no experience with proofs whatsoever. I understand the solution but I want to know if any of my first three tries have any correctness in them. Or is the solution above the only possible way to prove this equality without any other 'techniques'?

 

[robphy]

Are there restrictions on a and b? e.g. only integer solutions.

 

[CynicusRex]

Are there restrictions on a and b? e.g. only integer solutions.

Yes, only integers. No irrational or complex numbers as the books hasn't touched that yet.

 

[PeroK]

Homework Statement

Prove that if a² + ab + b² = 0 then a = 0 and b = 0
Hint: Recall the factorization of a³-b³. (Another solution will be discussed later when speaking about quadratic equations.)

Homework Equations

a² + ab + b² is close to a² + 2ab + b² = (a+b)²
a³-b³=(a-b)(a²+ab+b²)

The Attempt at a Solution

a² + ab + b² = 0
a² + ab + b² + ab = ab
a² + 2ab + b² = ab
(a+b)² = ab
(a+b)² - ab = 0

Here I figured, to get zero after a subtraction, both terms need to be equal. An addition of two numbers is the same to the multiplication of those two numbers only when a,b = 0 or a,b = 2. (0+0=0*0 and 2+2=2*2)
But since the first term is squared, (2+2)² - 2*2 = 16 - 4 ≠0. This leaves a,b = 0 as the only option to get a zero.

A second try was
a² + ab + b² = 0
a² + ab + b² + ab = ab
a² + 2ab + b² = ab
(a+b)² = ab
(a+b)² - ab = 0
(b+b)² - bb = 0 (given that a=b, and same works for b=a)
(2b)² - b² = 0
(2b-b)(2b+b) = 0
b*b = 0
b = 0

A third unsuccessful attempt
a² + ab + b² = 0
a² + ab + b² + ab = ab
a² + 2ab + b² = ab
(a+b)² = ab
a+b = √(ab)
a+b = √a√b
(a+b) - (√a√b) = 0

The solution:
a³-b³=(a-b)(a²+ab+b²), if (a²+ab+b²)=0, then a³-b³=0 or
a³ = b³ which can only happen when a=b. If a=b, then 0=a²+ab+b²=3b² which implies that
b=0 and a=b=0.

I have no experience with proofs whatsoever. I understand the solution but I want to know if any of my first three tries have any correctness in them. Or is the solution above the only possible way to prove this equality without any other 'techniques'?

Your proofs only get so far and then fail on the key step. An alternative would be to start with simply:

$a^2 + b^2 = -ab$

$a^2 + b^2 = |a||b|$

Can you justify that second step?

Then consider the case $|a| \ge |b|$

That's perhaps a more direct approach.

 

[PeroK]

Yes, only integers. No irrational or complex numbers as the books hasn't touched that yet.

You should be trying to prove it for any real numbers.

 

[CynicusRex]

Your proofs only get so far and then fail on the key step. An alternative would be to start with simply:

$a^2 + b^2 = -ab$

$a^2 + b^2 = |a||b|$

Can you justify that second step?

Then consider the case $|a| \ge |b|$

That's perhaps a more direct approach.

$a^2 + b^2 = -ab$

$a^2 + b^2 = |a||b|$

The second step can be justified by the fact that any number to an even power will be positive, except for zero.

$|a| \ge |b|$ With this I think you want to show me that, if one number is bigger than the other, there is no possibility that both numbers are zero, which is the only case that leads to the equation equaling zero.
So, if only one number would be zero, let's say a=0, then you're left with 0² + b² = |0||b| = 0, which can only mean b=0.
The other case, where both numbers are the same, results in:
a²+b² = |a||b|
a²+a² = |a||a|
2a² = a²
2a²-a² = 0
a² = 0
a = 0

Is this what you meant?

(Another side question: let's say I make the mistake of doing
2a² = a²
2 = a²/a² (can't divide by 0)
2 = 1
How's this called in mathematical terminology? )

 

[Buffu]

a² + ab + b² = 0

Another approach would be solving the quadratic for a. Actually you only need to consider the determinant to get the answer.

 

[CynicusRex]

Another approach would be solving the quadratic for a. Actually you only need to consider the determinant to get the answer.

Can not be used yet.

 

[Buffu]

Can not be used yet.

Is that a requirement by the author or you simply don't want ?Anyhow book's solution is ideal.
By the way what you are doing in your other attempts is what's called 'Completing the square' which is used to solve quadratic equations and quadratic formula is also derived by that method.

 

[PeroK]

$a^2 + b^2 = -ab$

$a^2 + b^2 = |a||b|$

The second step can be justified by the fact that any number to an even power will be positive, except for zero.

$|a| \ge |b|$ With this I think you want to show me that, if one number is bigger than the other, there is no possibility that both numbers are zero, which is the only case that leads to the equation equaling zero.
So, if only one number would be zero, let's say a=0, then you're left with 0² + b² = |0||b| = 0, which can only mean b=0.
The other case, where both numbers are the same, results in:
a²+b² = |a||b|
a²+a² = |a||a|
2a² = a²
2a²-a² = 0
a² = 0
a = 0

Is this what you meant?

(Another side question: let's say I make the mistake of doing
2a² = a²
2 = a²/a² (can't divide by 0)
2 = 1
How's this called in mathematical terminology? )

You've a bit of work to do to understand the logic of proofs. Often you are using what you are trying to prove during steps in the process.

For this proof there are two distinct approaches:

The direct approach where you assume nothing extra about $a, b$.

Proof by contradiction, where you assume one or both of $a, b$ are non zero and show this leads to a contradiction.

In either case, you need a clear strategy for your proof.

For this proof, the key thing you should be thinking is: Isn't $a^2 + b^2$ always bigger than $ab$? If $a$ is bigger than $b$ then $a^2$ on its own is bigger than $ab$.

Can you use that idea to guide you through a formal proof?

By the way, the simplest way to justify that second step was just to notice that the left hand side is non-negative, hence $-ab$ is non-negative, hence $-ab = |-ab| = |a||b|$.

 

[CynicusRex]

By the way, the simplest way to justify that second step was just to notice that the left hand side is non-negative, hence $-ab$ is non-negative, hence $-ab = |-ab| = |a||b|$.

Isn't that what I said? "The second step can be justified by the fact that any number to an even power will be positive, except for zero."
Meaning that the left hand side is positive and therefore also the right hand side.

I'll see what I can do for the formal proof. Thank you.

 

[CynicusRex]

You've a bit of work to do to understand the logic of proofs. Often you are using what you are trying to prove during steps in the process.

For this proof there are two distinct approaches:

The direct approach where you assume nothing extra about $a, b$.

Proof by contradiction, where you assume one or both of $a, b$ are non zero and show this leads to a contradiction.

In either case, you need a clear strategy for your proof.

For this proof, the key thing you should be thinking is: Isn't $a^2 + b^2$ always bigger than $ab$? If $a$ is bigger than $b$ then $a^2$ on its own is bigger than $ab$.

Can you use that idea to guide you through a formal proof?

By the way, the simplest way to justify that second step was just to notice that the left hand side is non-negative, hence $-ab$ is non-negative, hence $-ab = |-ab| = |a||b|$.

Okay so.
a²+ab+b² = 0
a²+b² = -ab
a²+b² = |a||b|

1. If a=b where a,b ≠ 0, then
a²+b² = |a||b| becomes
b²+b² = |b||b|
2b² = b² which is impossible unless when b=0, thus a=0 too.

2. If a>b where a,b ≠ 0, then
a²+b² = |a||b| becomes
a²+b² > |a||b| because a² > |a||b| and therefore a²+b² - |a||b| > 0
Going back to a²+ab+b² = 0
If a²+b² is greater than |a||b| then a²+ab+b² shouldn't be able to equal 0 because in either case when ab is positive or negative, it can not cancel a²+b² because the absolute value of ab is not large enough? So a>b is impossible which leaves a=b, which in turn leaves b=0 and a=0.

 

[Ray Vickson]

Can not be used yet.

Who told you that you cannot use the method of solving a quadratic equation?

 

[SammyS]

Who told you that you cannot use the method of solving a quadratic equation?

Look at the statement of the problem in OP.

Homework Statement

Prove that if a² + ab + b² = 0 then a = 0 and b = 0
Hint: Recall the factorization of a³-b³. (Another solution will be discussed later when speaking about quadratic equations.)

 

[PeroK]

Okay so.
a²+ab+b² = 0
a²+b² = -ab
a²+b² = |a||b|

1. If a=b where a,b ≠ 0, then
a²+b² = |a||b| becomes
b²+b² = |b||b|
2b² = b² which is impossible unless when b=0, thus a=0 too.

2. If a>b where a,b ≠ 0, then
a²+b² = |a||b| becomes
a²+b² > |a||b| because a² > |a||b| and therefore a²+b² - |a||b| > 0
Going back to a²+ab+b² = 0
If a²+b² is greater than |a||b| then a²+ab+b² shouldn't be able to equal 0 because in either case when ab is positive or negative, it can not cancel a²+b² because the absolute value of ab is not large enough? So a>b is impossible which leaves a=b, which in turn leaves b=0 and a=0.

I was thinking of a much simpler proof. Assume, without loss of generality, that $|a| \ge |b|$. Then:

$a^2 \ge |a||b|$

And, unless $b =0$:

$a^2 +b^2 > |a||b|$

That's the core of the proof. The key here is to use the fact that one of $a^2$ or $b^2$ on its own is larger than |ab|. With equality only when $a=b$.

 

[CynicusRex]

I was thinking of a much simpler proof. Assume, without loss of generality, that $|a| \ge |b|$. Then:

$a^2 \ge |a||b|$

And, unless $b =0$:

$a^2 +b^2 > |a||b|$

That's the core of the proof. The key here is to use the fact that one of $a^2$ or $b^2$ on its own is larger than |ab|. With equality only when $a=b$.

"And, unless $b = 0$"
Isn't the unless $b = 0$ redundant? Because the expression holds even when b = 0.
a² + 0² > a * 0. Or maybe you meant to say unless a,b = 0? Or is it added so the complete expression remains?

"I was thinking of a much simpler proof." Does that mean my proof is correct also? I'm self studying and have never learned how to write proofs yet, so I have nothing to go on. Thanks for your patience.

 

[PeroK]

"And, unless $b = 0$"
Isn't the unless $b = 0$ redundant? Because the expression holds even when b = 0.
a² + 0² > a * 0. Or maybe you meant to say unless a,b = 0? Or is it added so the complete expression remains?

"I was thinking of a much simpler proof." Does that mean my proof is correct also? I'm self studying and have never learned how to write proofs yet, so I have nothing to go on. Thanks for your patience.

Your proof goes round the houses, as it were. You get to the front door, the you run round the house 2-3 time and climb in a back window.

Also, you say one equation "becomes" another, that isn't logical.

You need to step back and decide what you are assuming, where this leads and how it proves the original statement.

Finally, my statement "unless $b =0$" was not redundant, as you need the strictly $>$ to rule out $b \ne 0$.

Do you want to have another go at a full proof? Here's the outline:

Assume $a^2 + ab + b^2 = 0$

Show that this implies

$a^2 + b^2 = |ab| \$ (1)

Assume, wlog (without loss of generality) that $|a| \ge |b|$

Show that $b = 0$, otherwise we have a contradiction to equation (1).

Show that also $a= 0$.

 

[lurflurf]

$$a^2+ab+b^2=\left.\left.\frac{1}{4}\right[3(a+b)^2+(a-b)^2\right]$$

 

[CynicusRex]

Your proof goes round the houses, as it were. You get to the front door, the you run round the house 2-3 time and climb in a back window.

Also, you say one equation "becomes" another, that isn't logical.

You need to step back and decide what you are assuming, where this leads and how it proves the original statement.

Finally, my statement "unless $b =0$" was not redundant, as you need the strictly $>$ to rule out $b \ne 0$.

Do you want to have another go at a full proof? Here's the outline:

Assume $a^2 + ab + b^2 = 0$

Show that this implies

$a^2 + b^2 = |ab| \$ (1)

Assume, wlog (without loss of generality) that $|a| \ge |b|$

Show that $b = 0$, otherwise we have a contradiction to equation (1).

Show that also $a= 0$.

K, I understand my redundancy mistake.
Isn't 'becomes' logical because it gives no information about the action you're performing?

I'll have another try at the proof after I've made some other exercises.

 

[abilolado]

if $a^2+ab+b^2=0$, assuming a and b are real, then
$a^2$ is positive
$b^2$ is positive
therefore $ab$ MUST be negative, in order for the sum to be 0.

The original equation can be rewritten as:
$(a+b)^2-ab=0$
Therefore $a+b=(ab)^{1/2}$
which the only real solution, if $ab$ IS negative, is ${a=0,b=0}$

 

[Ray Vickson]

if $a^2+ab+b^2=0$, assuming a and b are real, then
$a^2$ is positive
$b^2$ is positive
therefore $ab$ MUST be negative, in order for the sum to be 0.

The original equation can be rewritten as:
$(a+b)^2-ab=0$
Therefore $a+b=(ab)^{1/2}$
which the only real solution, if $ab$ IS negative, is ${a=0,b=0}$

No. If $ab < 0$ you cannot take its square root (at least, not in the real numbers), and of course, if $ab < 0$ you cannot have $a = b = 0$ (so you should never, ever write that). However, you can get a contradiction another way:
(1) You have just shown that $ab = -a^2 - b^2$, which is $\leq 0$.
(2) You can write $a^2 + b^2 + ab$ as $(a+b)^2 - ab$, so you also have $ab = (a+b)^2$, which is $\geq 0$.
(1) and (2) are compatible only when $a = b = 0$.

 

[abilolado]

No. If $ab < 0$ you cannot take its square root (at least, not in the real numbers), and of course, if $ab < 0$ you cannot have $a = b = 0$ (so you should never, ever wirite that). However, you can get a contradiction another way:
(1) You have just shown that $ab = -a^2 - b^2$, which is $\leq 0$.
(2) You can write $a^2 + b^2 + ab$ as $(a+b)^2 - ab$, so you also have $ab = (a+b)^2$, which is $\geq 0$.
(1) and (2) are compatible only when $a = b = 0$.

I should rephrase some stuff in my post, when I say must be positive or must be negative, I mean, $≥ 0 or ≤ 0$
So, $ab ≤ 0$
If $a+b=(ab)^{1/2}$
The only real solution is if $ab = 0$, exactly because there is no real root of a negative number, which means that $a=0$ or $b=0$ or $a=b=0$
if $ab = 0$, then $a+b=0$, so $a=-b$
Since one of them is 0, so must be the other.

 

[PeroK]

if $a^2+ab+b^2=0$, assuming a and b are real, then
$a^2$ is positive
$b^2$ is positive
therefore $ab$ MUST be negative, in order for the sum to be 0.

The original equation can be rewritten as:
$(a+b)^2-ab=0$
Therefore $a+b=(ab)^{1/2}$
which the only real solution, if $ab$ IS negative, is ${a=0,b=0}$

That's a neat approach. A better way to finish it off might be to say:

$a^2 + b^2 = -(a+b)^2$

And then $a=b=0$.

As already pointed out you should be saying non-positive instead of negative, as the latter excludes $0$.

 

[LCKurtz]

if $a^2+ab+b^2=0$, assuming a and b are real, then
$a^2$ is positive
$b^2$ is positive
therefore $ab$ MUST be negative, in order for the sum to be 0.

The original equation can be rewritten as:
$(a+b)^2-ab=0$
Therefore $a+b=(ab)^{1/2}$
which the only real solution, if $ab$ IS negative, is ${a=0,b=0}$

You have $(a+b)^2=ab$. Put that in the original equation $a^2+ab+b^2 = 0$ giving $a^2 + (a+b)^2 + b^2 = 0$ so $a=b=0$.

Edit: Only after posting this did I notice there was a second page of replies. But, hey, it's New Year's Eve, so why am I in here posting anyway.

 

[zexxa]

Well late to the party but I've been thinking about extending the domain of this question to including all numbers including complex numbers. This is my proof for it and I'd like to know if it is valid.

Given $a^2+ab+b^2=0$ ∀ $a,b ∈ ℂ$
$a^2+b^2=-ab$ ⇒(Realise that for this to be true only either $a,b∈ℝ$ or both $a,b∉ℝ$)$$ab = \begin{cases} ≤0 & \text{if } a,b ∈ ℝ \\ ≥0 & \text{if } a,b ∉ ℝ \end{cases}$$
$(a+b)^2-2ab=-ab$
$(a+b)^2=ab$
Suppose $a,b≠0$
$⇒(a+b)^2/(ab)=1$
Since $(a+b)^2$ and $ab$ have to be of the same sign, it is a contradiction with the case statement above and hence a,b cannot not be 0.

 

[lurflurf]

^complex numbers can have sign other than +1 and -1.
suppose a=r b
then
0=a^2+ab+b^2=r^2 b^2+r b^2+b^2=(1+r+r^2)b^2
1+r+r^2=0 or b^2=0
if r is real 1+r+r^2 cannot be zero
if r is complex it can be zero

 

[zexxa]

@lurflurf aha you're right. I was thinking only about imaginary numbers when I wrote it out and forgot that the variables doesn't have to only be real or imaginary numbers! Thanks for the correction!

 

[TeethWhitener]

if $a^2+ab+b^2=0$, assuming a and b are real, then
$a^2$ is positive
$b^2$ is positive
therefore $ab$ MUST be negative, in order for the sum to be 0.

The original equation can be rewritten as:
$(a+b)^2-ab=0$
Therefore $a+b=(ab)^{1/2}$
which the only real solution, if $ab$ IS negative, is ${a=0,b=0}$

$a^2$ is positive OR ZERO.
$b^2$ is positive OR ZERO.
therefore $ab$ MUST be negative OR ZERO, in order for the sum to be 0.
I'd point out too that $(a+b)^2$ is positive (or zero), which implies by $(a+b)^2-ab=0$ that $ab$ MUST be positive (or zero) in order for the sum to be zero. This implies that if $ab$ is nonzero, it must be positive and negative at the same time, which is impossible.

 

[CynicusRex]

I was thinking of a much simpler proof. Assume, without loss of generality, that $|a| \ge |b|$. Then:

$a^2 \ge |a||b|$

And, unless $b =0$:

$a^2 +b^2 > |a||b|$

That's the core of the proof. The key here is to use the fact that one of $a^2$ or $b^2$ on its own is larger than |ab|. With equality only when $a=b$.

I'm going through the same book again. You already pointed me in the right direction before, but think I got it now. However, I could be wrong.

$$\text{Prove if}\ a^{2}+ab+b^{2}={0} \ \text{, then}\ a=0 \ \text{and } b=0 \\ a^{2}+ab+b^{2}=0 \\ a^{2}+b^{2}=-ab$$
a², b² are always positive, so their sum must be positive. It follows, if a and b have the same sign, the right hand side is negative,
$$(-a)^{2}+(-b)^{2}=-(-a)(-b) \\ a^{2}+b^{2}=-ab$$ so the equation only holds when a=b=0.

If a and b have opposite signs, then a>b or b>a, and ab becomes positive,
$$a^{2}+(-b)^{2}=-a(-b) \\ a^{2}+b^{2}=|a||b|$$
so the equation can only hold when a=b=0 because a² + b² (and either a² or b² seperately as well) will otherwise always be larger than |ab|.

 

[PeroK]

I'm going through the same book again. You already pointed me in the right direction before, but think I got it now. However, I could be wrong.

$$\text{Prove if}\ a^{2}+ab+b^{2}={0} \ \text{, then}\ a=0 \ \text{and } b=0 \\ a^{2}+ab+b^{2}=0 \\ a^{2}+b^{2}=-ab$$
a², b² are always positive, so their sum must be positive. It follows, if a and b have the same sign, the right hand side is negative,
$$(-a)^{2}+(-b)^{2}=-(-a)(-b) \\ a^{2}+b^{2}=-ab$$ so the equation only holds when a=b=0.

If a and b have opposite signs, then a>b or b>a, and ab becomes positive,
$$a^{2}+(-b)^{2}=-a(-b) \\ a^{2}+b^{2}=|a||b|$$
so the equation can only hold when a=b=0 because a² + b² (and either a² or b² seperately as well) will otherwise always be larger than |ab|.

I don't follow your proof, I'm sorry to say.

 

[Mark44]

Try a proof by contradiction. The original statement is this: If $a^2 + ab + b^2 = 0$, then a = 0 and b = 0.
Suppose the hypothesis is true, but the conclusion is false, i.e., that If $a^2 + ab + b^2 = 0$, then it's not true that a = 0 and b = 0.
The negation of the conclusion is equivalent to $a \ne 0 \text{ or } b \ne 0$, by De Morgan's law.
The assumptions that $a \ne 0$ or $b \ne 0$ can be handled in four cases.
1. a < 0, b ≥ 0
$a^2 + ab + b^2 = 0 \Leftrightarrow a^2 + 2ab + b^2 = ab \Leftrightarrow (a + b)^2 = ab$
Show that this assumption leads to a contradiction.
2. a > 0, b ≥ 0
$a^2 + ab + b^2 = 0 \Leftrightarrow a^2 + b^2 = -ab$
Show that this assumption also leads to a contradiction
Cases 3 and 4 are similar, but in these cases, we make assumptions about b being negative or nonnegative.

By the way, in your work you seem to think that a and -a must necessarily be positive and negative, respectively. That's not true. If a = -3, then a < 0 while -a > 0.

 

[George Jones]

$$0=a^2 + ab + b^2$$

Complete the square for the first two terms on the right side.

 

[Best Pokemon]

You can use a proof by contrapositive. Your statement is of the form if $P$ then $Q$. This is a conditional statement. The contrapositive is: if $not~ Q$ then $not~ P$. Conditional statements are equivalent to their contrapositives.
$Q$ is of the form ($A$ and $B$) where $A$ is $a=0$ and $B$ is $b=0$. $not~ Q$ is $not~(A$ and $B)$ which is also ($not~ A$ or $not~ B$) according to De Morgan's laws.
So $not~ Q$ is $a\neq0$ or $b\neq0$
NB $not~ A$ is $a\neq0$ since the negation of $a$ equals $0$ is $a$ is not equal to $0$. You do the same for $not~ B$, and $not~ P$

Proof:
You can rewrite the statement as: if $a\neq0$ or $b\neq0$ then $a^2 +ab +b^2 \neq0$
Now $a\neq0$ or $b\neq0$ means at least one of them is non-zero. So if you pick one to be $0$ and the other to be a non-zero number $c$ you would get:
$c^2 + 0c + 0^2$
$=c^2$ which is a non-zero number.

So this proves it.

 

[StoneTemplePython]

I suppose it's a matter of taste, but I prefer to bound things over most 'contra' proofs.

Via $GM \leq AM$ or Cauchy's Inequality or playing around with $(a-b)^2 \geq 0$, while dealing in $\mathbb R$, we know real numbers squared are always non-negative. We put this together and get the below bound

$a^2 + b^2 = \big \vert a^2\big \vert + \big \vert b^2 \big \vert = \big \vert a^2 + b^2 \big \vert = \big \vert ab \big \vert \leq \big \vert \frac{1}{2}\big(a^2 + b^2\big) \big \vert = \frac{1}{2}\big(\big \vert a^2\big \vert + \big \vert b^2 \big \vert\big)= \frac{1}{2}\big(a^2 + b^2\big)$

subtract $\frac{1}{2}\big(a^2 + b^2\big)$ from each side and get

$\frac{1}{2}\big(a^2 + b^2\big) \leq 0$

and if the sum of two real non-negative numbers is zero, then both must be zero. (A fancier way of saying this would refer to positive definiteness but it's not really needed.)

 

[CynicusRex]

A year ago I thought I'd never become good at skateboarding because I kept failing at one simple trick. After lots of practice, something clicked this month, and things start to feel natural. I really hope this happens to proofs \(\displaystyle as well, because I'm sure as hell feeling quite pathetic at the moment.\)