Proving |a-b| ≤ $\sqrt{a^2 + b^2}$ Using Induction

[knowLittle]

Homework Statement

Prove for any $a,b \in \mathbb{R^+} : |a-b| \leq \sqrt{a^2 +b^2}$

The Attempt at a Solution

How should I start? Can I use induction?
Should I use contrapositive?

 

[Curious3141]

Homework Statement

Prove for any $a,b \in \mathbb{R^+} : |a-b| \leq \sqrt{a^2 +b^2}$

The Attempt at a Solution

How should I start? Can I use induction?
Should I use contrapositive?

You're overthinking it. Consider $(|a-b|)^2$ first.

 

[knowLittle]

uh, are you serious?

$(|a-b|)^2 = (a^2 - 2ab + b^2 ) \leq (\sqrt{a^2 +b^2})^2 = a^2 +b^2$

Then the LHS will always be lesser for any $\mathbb{R^+}$

 

[Curious3141]

uh, are you serious?

$(|a-b|)^2 = (a^2 - 2ab + b^2 ) \leq (\sqrt{a^2 +b^2})^2 = a^2 +b^2$

Then the LHS will always be lesser for any $\mathbb{R^+}$

Pretty much, except that going from this step to this: $(a^2 - 2ab + b^2 ) \leq (\sqrt{a^2 +b^2})^2$ is unnecessary. You can just go from this to this: $(a^2 - 2ab + b^2 ) \leq a^2 +b^2$ by observing that $2ab > 0$.

As a final step, you still have to take the square root of both sides. You have to justify that the inequality doesn't change direction by observing the monotonic increasing nature of the square root function (just a simple sketch would do).

BTW, if a and b are both positive reals, the strict inequality sign can be used (a stronger statement), i.e. $|a-b| < \sqrt{a^2 + b^2}.$

 

[knowLittle]

Wouldn't I need this first:

$-\sqrt{ a^2 + b^2} < (a-b) < \sqrt{ a^2 +b^2 }\\$

Then,
$a^2 +b^2 < ( a-b )^2 = (a^2 - 2ab + b^2 ) < (\sqrt{a^2 +b^2})^2 = a^2 +b^2$
Now, I don't understand

 

[Curious3141]

Wouldn't I need this first:

$-\sqrt{ a^2 + b^2} < (a-b) < \sqrt{ a^2 +b^2 }\\$

Then,
$a^2 +b^2 < ( a-b )^2 = (a^2 - 2ab + b^2 ) < (\sqrt{a^2 +b^2})^2 = a^2 +b^2$
Now, I don't understand

I'm sorry, I do not understand what you're doing here.

The first step is to consider the square of the expression $|a-b|$.

It should be trivial to recognise that $(|a-b|)^2 = |(a-b)^2| = (a-b)^2 = a^2 + b^2 - 2ab$

The next step is to note that $a^2 + b^2 - 2ab < a^2 + b^2$ for positive reals a and b.

That gives you $(|a-b|)^2 < a^2 + b^2$. The final step is to take the square root of both sides.

Can you finish it off from here?

 

[knowLittle]

I'm sorry, I do not understand what you're doing here.

The first step is to consider the square of the expression $|a-b|$.

It should be trivial to recognise that $(|a-b|)^2 = |(a-b)^2| = a^2 + b^2 - 2ab$

The next step is to note that $a^2 + b^2 - 2ab < a^2 + b^2$ for positive reals a and b.

That gives you $(|a-b|)^2 < a^2 + b^2$. The final step is to take the square root of both sides.

Can you finish it off from here?

I see what you are doing.

About what I wrote prior to this...
I was talking about this supposedly property of absolute value
$|a-b| < c$, then
$-c< a-b <c$

 

[Curious3141]

I see what you are doing.

About what I wrote prior to this...
I was talking about this supposedly property of absolute value
$|a-b| < c$, then
$-c< a-b <c$

That is true, but not relevant here.