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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  • $\begingroup$ This is the AM-QM inequality applied to $\sqrt{a}$ and $\sqrt{b}$. $\endgroup$ – Element118 Nov 24 '15 at 11:50
  • 1
    $\begingroup$ ... or the concavity of the square root function ... or ... $\endgroup$ – mickep Nov 24 '15 at 11:57
  • $\begingroup$ Possible duplicate of Arithmetic mean <= Quadratic mean, proof? $\endgroup$ – user147263 Nov 24 '15 at 23:34

$‎\left( ‎\frac{\sqrt{a}+ \sqrt{b} }{2} ‎\right) ‎^{2}‎‎ \leq ‎\left( ‎\sqrt{ \frac{a+b}{2} }‎\right) ‎^{2}‎‎$ ‎$ ‎\Longleftrightarrow‎ ‎‎\left( ‎\sqrt{a}‎‎‎-‎\sqrt{b}‎‎\right)‎^{2}‎ ‎‎\geqslant ‎0‎ $‎

  • $\begingroup$ What we need here is a $\Leftarrow$ or $\Leftrightarrow$, not a $\Rightarrow$. $\endgroup$ – Wojowu Nov 24 '15 at 13:07

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*}


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