A four-variable maximization problem We let function
\begin{equation}
\begin{aligned}
f(x_1,~x_2,~x_3,~x_4) ~&=~ \sqrt{(x_1+x_2)(x_1+x_3)(x_1+x_4)} \\
        &+ \sqrt{(x_2+x_1)(x_2+x_3)(x_2+x_4)} \\
        &+ \sqrt{(x_3+x_1)(x_3+x_2)(x_3+x_4)} \\
        &+ \sqrt{(x_4+x_1)(x_4+x_2)(x_4+x_3)},
\end{aligned}
\end{equation}
where variables $x_1,~x_2,~x_3,~x_4$ are positive and satisfy $x_1+x_2+x_3+x_4 ~=~ 1$. We want to prove that $f(x_1,~x_2,~x_3,~x_4)$ attains its global maximum when $x_1=x_2=x_3=x_4=\frac{1}{4}$.
This looks a difficult problem even if it is at high-school level. Any clues? Your ideas are greatly appreciated.
 A: Using the inequality of the means for 2 variables, $(\sqrt{ab}\leq\frac{a+b}{2})$ for positive $a$ and $b$, equality only when $a=b$ we have
\begin{equation}
\begin{aligned}
&\sqrt{(x_1+x_2)(x_1+x_3)(x_1+x_4)} + \sqrt{(x_2+x_1)(x_2+x_3)(x_2+x_4)} \\&=\sqrt{(x_1+x_2)}(\sqrt{(x_1+x_3)(x_1+x_4)}+\sqrt{(x_2+x_3)(x_2+x_4)})\\ &\leq \sqrt{(x_1+x_2)}(\frac{(x_1+x_3)+(x_1+x_4)}{2}+\frac{(x_2+x_3)+(x_2+x_4)}{2})\\&=\sqrt{(x_1+x_2)}(x_1+x_2+x_3+x_4)\\&=\sqrt{(x_1+x_2)}
\end{aligned}
\end{equation}
with equality attained only when $x_3=x_4$.
Similarly or by swapping $x_1$ and $x_3$, $x_2$ and $x_4$ we have
\begin{equation}
\begin{aligned}
&\sqrt{(x_3+x_1)(x_3+x_2)(x_3+x_4)} + \sqrt{(x_4+x_1)(x_4+x_2)(x_4+x_3)} \\&\leq\sqrt{(x_3+x_4)}
\end{aligned}
\end{equation}
with equality attained only when $x_1=x_2$.
Adding the two inequalities we obtain
\begin{equation}
\begin{aligned}
f(x_1,~x_2,~x_3,~x_4) ~&\leq \sqrt{(x_1+x_2)}+\sqrt{(x_3+x_4)}
\end{aligned}
\end{equation}
By Jensen's Inequality since $-\sqrt{x}$ is convex for $0\leq x \leq 1$ we have 
\begin{equation}
\begin{aligned}
\frac{\sqrt{(x_1+x_2)}+\sqrt{(x_3+x_4)}}{2}\leq \sqrt{\frac{(x_1+x_2)+(x_3+x_4)}{2}}=\sqrt{\frac{1}{2}}
\end{aligned}
\end{equation}
with equality attained only when $(x_1+x_2)=(x_3+x_4)=\frac{1}{2}$.
Hence 
\begin{equation}
\begin{aligned}
f(x_1,~x_2,~x_3,~x_4) ~&\leq \sqrt{2}
\end{aligned}
\end{equation}
with equality only attained if $x_1=x_2$, $x_3=x_4$ and $(x_1+x_2)=(x_3+x_4)=\frac{1}{2}$ which is equivalent to $x_1=x_2=x_3=x_4=\frac{1}{4}.$
A: By Cauchy-Schwarz for sums (with $x_i=\sqrt{a+b}$ and $y_i=\sqrt{a+c}\sqrt{a+d}$, also, I am using cyclic sum notation):
\begin{split}
f(a,b,c,d)&=\sum_{\text{cyc}} \sqrt{(a+b)(a+c)(a+d)}\\&\le \left(\left(\sum_{\text{cyc}} a+b\right) \cdot \left(\sum_{\text{cyc}} (a+c)\cdot (a+d)\right)\right)^{\frac12}\\
&=\sqrt{\big(2(a+b+c+d)\big)\cdot\big((a+b+c+d)^2\big)}\\
&=\sqrt{2}.
\end{split}
Equality occurs if and only if $(a+b,b+c,c+d,d+a)$ is a constant multiple of $((a+c)\cdot (a+d),\dots)$  which, together with $a+b+c+d=1$ implies $a=b=c=d=\frac14$.
