Could anyone help me to proof this inequality: $\frac{\sqrt{a}+ \sqrt{b} }{2} \leq \sqrt{ \frac{a+b}{2} }$ for $a \geq 0$ and $b \geq 0$.
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This is the AM-QM inequality applied to $\sqrt{a}$ and $\sqrt{b}$. – Element118 Nov 24 '15 at 11:50
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1... or the concavity of the square root function ... or ... – mickep Nov 24 '15 at 11:57
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Possible duplicate of Arithmetic mean <= Quadratic mean, proof? – Nov 24 '15 at 23:34
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Note that \begin{align*} (\sqrt{a}-\sqrt{b})^2 \ge 0 &\implies a+b \ge 2\sqrt{ab} \\ &\implies \frac{a+b}{2} \ge \sqrt{ab} \\ &\implies a+b \ge \frac{a+b}{2}+\sqrt{ab}\\ &\implies \frac{a+b}{2} \ge \frac{a+b+2\sqrt{ab}}{4} \\ &\implies \frac{a+b}{2} \ge\left(\frac{\sqrt{a}+\sqrt{b}}{2}\right)^2\\ &\implies \sqrt{\frac{a+b}{2}} \ge \frac{\sqrt{a}+\sqrt{b}}{2} \end{align*}
OrangeApple3
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$\left( \frac{\sqrt{a}+ \sqrt{b} }{2} \right) ^{2} \leq \left( \sqrt{ \frac{a+b}{2} }\right) ^{2}$ $ \Longleftrightarrow \left( \sqrt{a}-\sqrt{b}\right)^{2} \geqslant 0 $
naser
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What we need here is a $\Leftarrow$ or $\Leftrightarrow$, not a $\Rightarrow$. – Wojowu Nov 24 '15 at 13:07