How to prove statement with quantifiers and inequalities?

Question

The statement goes as follows:

$ \forall a, b > 0, a + b \geq 2\sqrt{ab} $

Which I am interpreting as "for all a and b greater than $0$, $a+b \geq 2\sqrt{ab}$".

I am now trying to reason through the logic.

So:

$a + b \geq 2\sqrt{ab}$

$ \implies a^2 + 2ab + b^2 \geq 4ab $

$ \implies a^2 + b^2 \geq 2ab $

Should I create a side proof proving that for all $ x,y > 0$, $x^2+y^2 > 2xy$. Thus, by this side proof, this is the case?

But how would I even prove this side proof?

Thanks in advance!

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I feel like my proof the way I would like to present it, is way too circular.

Answer 1

If $a,b>0$, then $a+b-2\sqrt{ab}=\sqrt a^2-2\sqrt a\sqrt b+\sqrt b^2=\bigl(\sqrt a-\sqrt b\bigr)^2\geqslant0$. So, $a+b\geqslant2\sqrt{ab}$.

Answer 2

Your proof is right!

Since $a>0$ and $b>0$, we obtain $$a+b\geq2\sqrt{ab}\Leftrightarrow (a+b)^2\geq4ab\Leftrightarrow(a-b)^2\geq0$$ and we are done!

The squaring is possible because $a$ and $b$ are non-negatives.

If it's not so, then we can not do it always:

$-2>-1$ is true, but $(-2)^2>(-1)^2$ is wrong;

$-2>3$ is true and $(-2)^2>3^2$ is true.

In another hand, if $x\geq y\geq0$ then: $$x\geq y\Leftrightarrow x^2\geq y^2,$$ which says that your first step gives the proof.