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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.

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  • $\begingroup$ MSE is a right forum for such type questions. $\endgroup$ – user64494 Nov 18 at 15:26
  • $\begingroup$ @user64494 Do you have any justification for repeating this comment other than your belief that "art for art's sake is not research" and that a CAS verdict/solution removes the need/interest for an "analytic proof"? $\endgroup$ – Yemon Choi Nov 19 at 10:06
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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 $x_1=x_2=x_3=x_4=\frac{1}{4}.$

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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 of and only of $(a+b,b+c,c+d,d+a)$ is a constant multiple of $((a+c)\cdot (a+d),\dots)$ which, together with $a+b+c+d=1$ implies $a=b=c=d=\frac14$.

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