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Let $a>0$ and $b>0$. Prove that $\sqrt{ab} \le (a+b)/2$.

Here is what I have tried: Let $a \le b$. Multiplying both sides of this inequality by $a$ results in $a^2 \le ab$. It follows that $a \le \sqrt{ab}$. We can do the same process with $b$ to find that $a \le \sqrt{ab} \le b$.

Similarly we can add $a$ to both sides, then divide by 2 to give this inequality: $a \le (a+b)/2 \le b$.

However, I have no way to compare $\sqrt{ab}$ and $(a+b)/2$.

Martin Sleziak
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user215930
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  • Take the square of both sizes and use the fact that $a^{2}+b^{2}\geq2ab $ (do you see why?). – Nicolas Feb 16 '15 at 18:53
  • See http://math.stackexchange.com/questions/543253/how-can-i-prove-frac2xyxy-leq-sqrtxy-leq-fracxy2, http://math.stackexchange.com/questions/1114615/if-0ab-prove-that-a-sqrtab-fracab2b, http://math.stackexchange.com/questions/904827/how-to-prove-that-fracab2-geq-sqrtab-for-a-b0 – Martin Sleziak Feb 16 '15 at 19:36
  • This is an overkill but you can prove AM-GM inequality – kingW3 Feb 16 '15 at 19:46

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Hint: Since squares are non-negative, you can use that $$ (\sqrt{a}-\sqrt{b})^2 \geq 0. $$

N.U.
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