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Prove that for any two positive real numbers $a,b$ we have $(ab)^{\frac{1}{2}} \leq \frac{a+b}{2}$ and moreover that equality occurs if and only if $a = b$.

My attempt at a solution (incomplete):

Suppose $(ab)^{\frac{1}{2}} \leq \frac{a+b}{2}$, then

$4ab \leq (a+b)^2$

$\Rightarrow 4ab \leq a^2 + 2ab + b^2$

$\Rightarrow 0 \leq a^2 - 2ab + b^2 \Rightarrow 0 \leq (a-b)^2$.

How would I come to the conclusion that $a = b$ here? Am I allowed to say that for $(a-b)^2$ to meet the requirement of being equal to zero, $a = b$?

Also, for the only if proof, would I just take the last statement of the if proof and reconstruct the original equation or do I prove it in some other way?

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    That wasn't me who edited that. – 2017-02-02
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    @Jan I fixed it. – 2017-02-02
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    @Fox you can use @(insert user name here) to ping users. – 2017-02-02

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Note that since you know $$(a-b)^2=0 \implies a-b=\sqrt{0}=0$$

Also, your proof of the inequality is finished, as you have proved$$\frac{a+b}{2} \ge \sqrt{ab} \iff (a-b)^2 \ge 0$$ And the inequality on the right is well known.

So $$\frac{a+b}{2} =\sqrt{ab} \implies a=b$$

Also $$a=b \implies \frac{a+b}{2}=a=\sqrt{ab}$$

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    I noticed that in your answer the less than symbol was omitted. May I ask why that is okay? – 2017-02-02
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    @Fox Yes it is in the second equation. The "iff" statement is saying we should prove $$\frac{a+b}{2}=\sqrt{ab} \iff a=b$$ So I am not omitting the less than symbol. – 2017-02-02