A quick way to prove the inequality $\frac{\sqrt x+\sqrt y}{2}\le \sqrt{\frac{x+y}{2}}$

Can anyone suggest a quick way to prove this inequality?

$$\frac{\sqrt x+\sqrt y}{2}\le \sqrt{\frac{x+y}{2}}$$

• For a proof of a more general case (full QM-AM inequality), see here. – user26486 Apr 25 '15 at 10:27

Both sides are positive so you may square the inequality ot obtain \begin{align*}\left(\frac{\sqrt{x}+\sqrt{y}}{2}\right)^2\le\left(\sqrt{\frac{x+y}{2}}\right)^2 &\iff \frac{x+2\sqrt{xy}+y}{4}\le \frac{x+y}{2} \\&\iff \frac{2\sqrt{xy}}{4}\le\frac{x+y}{2}-\frac{x+y}{4} \\&\iff \sqrt{xy}\le \frac{x+y}{2}\\[0.2cm]& \iff (\sqrt{x}-\sqrt{y})^2\ge 0\end{align*} which is true.

• The last $\iff$ seems off to me. – GFauxPas Mar 23 '15 at 21:31
• it was showing the pre-edit answer to me, nevermind! – GFauxPas Mar 23 '15 at 21:38
• @GFauxPas Due to your comment I found the mistake! You were correct, thanks for noticing! – Jimmy R. Mar 23 '15 at 21:38
• Citing AM-GM is a bit overkill here. $\sqrt{xy}\le\frac{x+y}{2}\iff (\sqrt{x}-\sqrt{y})^2\ge 0$. AM-GM refers to $\sqrt[n]{x_1x_2\cdots x_n}\le \frac{x_1+x_2+\cdots+x_n}{n}$, $n\in\mathbb Z^+, x_1,\ldots,x_n >0$, which is difficult to prove. – user26486 Mar 23 '15 at 23:53
• @user31415 I agree. I always prefer to just note $a^2 - 2ab + b^2 \geq 0$ when dealing with AM-GM when $n = 2$. – MCT Mar 24 '15 at 0:48

$2ab \leq a^2+b^2$ so $\frac{a+b}{2} \leq \sqrt{\frac{a^2+b^2}{2}}$. Now choose $a = \sqrt{x},b = \sqrt{y}$.

• Just square the inequality $\frac{a+b}{2} \leq \sqrt{\frac{a^2+b^2}{2}}$. It is one of the most basic inequalities. All mean-type inequalities reduce to $(a-b)^2 \geq 0$. – Beni Bogosel Mar 23 '15 at 21:11
• @Stef: To such a question (no background, no reason, no details) I did not want to give an explicit answer. Anyone who stared more than two minutes at the inequality would square both sides, and start reducing what can be reduced. – Beni Bogosel Mar 23 '15 at 21:23

By Jensen's Inequality, $$\frac{1}{n} \sum \phi(x_i) \geq \phi \left(\frac{1}{n} \sum x_i \right)$$ for $\phi(x)$ convex ($\leq$ for $\phi$ concave). In your case, $\phi(x) = \sqrt{x}$ has a negative second derivative and is hence concave, so with $n=2$ the inequality immediately holds using the concave case.

assuming $$x\geq0 \land y\geq0$$

$$\dfrac{\sqrt{x}+\sqrt{y}}{2} \leq \sqrt{\dfrac{x+y}{2}}$$

Squaring:

$$\dfrac{x+y+2\sqrt{xy}}{4} \leq \dfrac{x+y}{2} \implies \dfrac{\sqrt{xy}}{2} \leq \dfrac{x+y}{4}$$

Squaring again

$$\dfrac{\sqrt{xy}}{2} \leq \dfrac{x+y}{4} \implies \dfrac{xy}{4} \leq \dfrac{x^2+2xy+y^2}{16}\implies$$

$$0 \leq \dfrac{x^2-2xy+y^2}{16} \implies 0 \leq \dfrac{(x-y)^2}{4}$$

But we now that $a^2\geq 0 \forall a \in \mathbb{R}$ so it's true...

• After squaring it's $2\sqrt{xy}$ not $\sqrt{2xy}$. – Fan Zheng Mar 24 '15 at 4:44
• Gahh, you're right. Edited :) – 3d0 Mar 24 '15 at 13:17

Your inequality is a special case ($n=2$) of QM-AM (quadratic-arithmetic mean) inequality: $$\sqrt{\frac{a_1^2+\cdots+a_n^2}{n}}\ge \frac{a_1+\cdots+a_n}{n}$$

In general, $a_i>0,k_2>k_1$ gives (Power Mean Inequality): $$\sqrt[k_2]{\frac{a_1^{k_2}+\cdots+a_n^{k_2}}{n}}\ge \sqrt[k_1]{\frac{a_1^{k_1}+\cdots+a_n^{k_1}}{n}}$$

This is a special case of Cauchy-Schwarz ($a_i,b_i>0$): $$\sqrt{a_1b_1}+\cdots+\sqrt{a_nb_n}\le\sqrt{(a_1+\cdots+a_n)(b_1+\cdots+b_n)}$$

$$\frac{\sqrt{x}+\sqrt{y}}{2}\le\sqrt{\frac{x+y}{2}}\iff\sqrt{x}+\sqrt{y}\le\sqrt{(1+1)(x+y)}$$

Divide by $\sqrt y$ and let $t=\frac xy$ to obtain $\frac{\sqrt t+1}2\leq\sqrt\frac{t+1}2$. Comparing the derivatives of both sides, we see $$\begin{array}{c}\left(\frac{\sqrt t+1}2\right)'&&\left(\sqrt\frac{t+1}2\right)'\\ \|&&\|\\ \frac1{2\sqrt{4t}}&\leq&\frac1{2\sqrt{2t+2}}&\quad\text{for }t\geq1\\ \frac1{2\sqrt{4t}}&\geq&\frac1{2\sqrt{2t+2}}&\quad\text{for }t\leq1\end{array}$$ so $$\frac{\sqrt t+1}2=-1+\int_1^u\left(\frac{\sqrt u+1}2\right)'du\leq-1+\int_1^u\left(\sqrt\frac{u+1}2\right)'du=\sqrt\frac{t+1}2$$ for all $t\geq0$.