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?
$$\frac{\sqrt x+\sqrt y}{2}\le \sqrt{\frac{x+y}{2}}$$
 A: 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.
A: $2ab \leq a^2+b^2$ so $\frac{a+b}{2} \leq \sqrt{\frac{a^2+b^2}{2}}$. Now choose $a = \sqrt{x},b = \sqrt{y}$.
A: 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...
A: 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.
A: This is a special case of Cauchy-Schwarz ($a_i,b_i>0$):  $$\sqrt{a_1b_1}+\cdots+\sqrt{a_nb_n}\le\sqrt{(a_1+\cdots+a_n)(b_1+\cdots+b_n)}$$  
$$\frac{\sqrt{x}+\sqrt{y}}{2}\le\sqrt{\frac{x+y}{2}}\iff\sqrt{x}+\sqrt{y}\le\sqrt{(1+1)(x+y)}$$
A: 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): $$\sqrt[k_2]{\frac{a_1^{k_2}+\cdots+a_n^{k_2}}{n}}\ge \sqrt[k_1]{\frac{a_1^{k_1}+\cdots+a_n^{k_1}}{n}}$$
A: 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 $t\geq0$.
