-1
$\begingroup$

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 $b \geq 0$.

$\endgroup$

closed as off-topic by Did, Davide Giraudo, Michael Albanese, user147263, Harish Chandra Rajpoot Nov 25 '15 at 0:08

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Davide Giraudo, Michael Albanese, Harish Chandra Rajpoot
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ What we need here is a $\Leftarrow$ or $\Leftrightarrow$, not a $\Rightarrow$. $\endgroup$ – Wojowu Nov 24 '15 at 13:07
4
$\begingroup$

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*}

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.