Sum of square roots inequality 
For all $a, b, c, d > 0$, prove that
$$2\sqrt{a+b+c+d} ≥ \sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}$$

The idea would be to use AM-GM, but $\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}$ is hard to expand. I also tried squaring both sides, but that hasn't worked either. Using two terms at a time doesn't really work as well. How can I solve this question? Any help is appreciated.
 A: You can use QM-AM for $\sqrt{a},\ \sqrt{b},\ \sqrt{c},\ \sqrt{d}$; I'm leaving the details for you.
A: If you square both sides:
$(\sqrt a + \sqrt b + \sqrt c +\sqrt d)^2 =a + b+c+d + 2(\sqrt{ab} + \sqrt{ac}+\sqrt{ad} + \sqrt{bc}+\sqrt{bd} + \sqrt{cd})$
while $(2\sqrt {a+b+c+d})^2= 4(a+b+c+d)$ so it suffices to prove
$2(\sqrt{ab} + \sqrt{ac}+\sqrt{ad} + \sqrt{bc}+\sqrt{bd} + \sqrt{cd})\le 3(a+b+c+d)$.
If we apply AM/GM though, we get $\sqrt{ab} \le \frac {a+b}2$ or in other words $2\sqrt{ab} \le a+b$.
And that does it:
$2(\sqrt{ab} + \sqrt{ac}+\sqrt{ad} + \sqrt{bc}+\sqrt{bd} + \sqrt{cd})\le$
$(a+b) + (a+c) + (a+d) + (b+c) + (b+d) = 3(a+b+c+d)$.
A: How about Jensen's inequality? $\sqrt x$ is concave so
$$\sqrt{\frac{a+b+c+d}{4}} \ge \frac 14 \left( \sqrt a +\sqrt b +\sqrt c +\sqrt d \right)$$
A: Alternative solution using Cauchy-Schwarz, which finishes the problem off immediately. By C-S, we have:
\begin{align}
& (a+b+c+d)(1+1+1+1) \geq (\sqrt{a}+ \sqrt{b}+ \sqrt{c}+ \sqrt{d} )^2 \\
& \Rightarrow 4(a+b+c+d) \geq (\sqrt{a}+ \sqrt{b}+ \sqrt{c}+ \sqrt{d} )^2  \\
& \Rightarrow 2\sqrt{a+b+c+d} \geq \sqrt{a}+ \sqrt{b}+ \sqrt{c}+ \sqrt{d}.
\end{align}
