Proof of \sqrt{ab}>\frac{2ab}{a+b} for Positive and Unequal Integers

[odolwa99]

Can anyone help me confirm if I've solved this correctly?

Many thanks.

Homework Statement

Prove that $\sqrt{ab}>\frac{2ab}{a+b}$ if a & b are positive & unequal.

Homework Equations

The Attempt at a Solution

if $(\sqrt{ab})^2>(\frac{2ab}{a+b})^2$
if $ab>\frac{4a^2b^2}{(a+b)^2}$
if $ab(a^2+2ab+b^2)>4a^2b^2$
if $a^3b+2a^2b^2+ab^3-4a^2b^2>0$
if $a^3b-2a^2b^2+ab^3>0$
if $(a^2b-ab^2)>0$...true

 

[Mark44]

Can anyone help me confirm if I've solved this correctly?

Many thanks.

Homework Statement

Prove that $\sqrt{ab}>\frac{2ab}{a+b}$ if a & b are positive & unequal.

Homework Equations

The Attempt at a Solution

if $(\sqrt{ab})^2>(\frac{2ab}{a+b})^2$

You are essentially starting off by assuming what you are to show.

if $ab>\frac{4a^2b^2}{(a+b)^2}$
if $ab(a^2+2ab+b^2)>4a^2b^2$
if $a^3b+2a^2b^2+ab^3-4a^2b^2>0$
if $a^3b-2a^2b^2+ab^3>0$
if $(a^2b-ab^2)>0$...true

Why? There's no guarantee that a²b - ab² > 0, so you can't say with any certainty that this is true.

Also, you should not be starting each statement with "if". The hypothesis (the "if" part, the part that you assume to be true) for this problem is this:
If a > 0 and b > 0 and a ≠ b

The conclusion (and what you need to prove) is this:
Then $\sqrt{ab}>\frac{2ab}{a+b}$

Each statement that you write should imply the truth of the statement that follows. The symbol to use is $\Rightarrow$.

One way to do this problem is a proof by contradiction. Assume that a > 0 and b > 0 and a ≠ b, AND that $\sqrt{ab} \leq \frac{2ab}{a+b}$

When you arrive at a contradiction (and you should), you will have proved the original statement.

 

[odolwa99]

I can see what you're getting at, but I've included an image of an example question from the textbook I'm using. Is the books approach wrong, or have I followed their method incorrectly?

 

[aralbrec]

I don't like that notation either. Implies is much better.

The very last step is not right. Trying factoring out ab and then notice you have a quadratic equation that can be factored. The proposition is false if you can find *any* a,b satisfying the conditions to make the inequality untrue.

 

[odolwa99]

Ok, thank you for checking that.

 

[SammyS]

...

if $a^3b-2a^2b^2+ab^3>0$
if $(a^2b-ab^2)>0$...true

$a^3b-2a^2b^2+ab^3\ne (a^2b-ab^2)$

Factor out ab from a³b-2a²b²+ab³ and you will get ab times a perfect square.

Regarding you textbook's example, which I think we've seen before:

It reminds me of the steps one might take on scratch paper to figure out how you might go about doing the actual proof. Starting with what you want to prove, you work backwards toward something(s) you know to be true.

So, for the case of the example:

Assuming a>0 and b>0,

$\displaystyle \frac{a}{b}+\frac{b}{a}\ge2\$ is true if $\displaystyle a^2+b^2\ge2ab\ .$

$\displaystyle a^2+b^2\ge2ab\$ is true if $\displaystyle a^2-2ab+b^2\ge0\ .$

$\displaystyle a^2-2ab+b^2\ge0\$ is equivalent to $\displaystyle (a-b)^2\ge0\$ which true.​

Most mathematicians would look at this proof as being written in reverse of the usual order.

IMO: The following is a more conventional order.

Assume a>0 and b>0.

$\displaystyle (a-b)^2\ge0\,,\$ the square of any real number is non-negative.

Expanding the left-hand side gives us: $\displaystyle a^2-2ab+b^2\ge0\ .$

Adding 2ab to both sides results in: $\displaystyle a^2+b^2\ge2ab\ .$

ab>0, so dividing both sides of the inequality by ab gives the desired result, namely: $\displaystyle \frac{a}{b}+\frac{b}{a}\ge2\ .$​

 

[odolwa99]

Thanks again, Sammy.

I was on the this site, a few weeks back, with questions about Proof by Induction, and there was a lot of disagreement then about how those questions should be answered. This book is actually meant for state exams, so I don't know if a drop in the quality of solution, compared to university level math, is a consequence of that, or just a matter of preference by the author of this particular text...?

Either way, thank you for the clarification.

 

[aralbrec]

The notation in the book is not incorrect; it's just not what most of us are used to seeing.

If..then is the logical statement implies.

If A then B

is the same as

A -> B

The statement is false only if A can be found to be true while B is false. This is exactly what the problem in the book is doing, based on a chain of "if A then B, if B then C,..."

 

[odolwa99]

I'll make note of this, too. Thanks.