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Post Closed as "Not suitable for this site" by user44191, J. M. isn't a mathematician, Alex M., ARG, Alexey Ustinov
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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 inat high-school level. Any clues? Your helpful ideas are greatly appreciated.

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 in high-school level. Any clues? Your helpful ideas are greatly appreciated.

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.

Used LaTeX to rewrite formulas
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We let function

$f(x_1,~x_2,~x_3,~x_4) ~=~ \sqrt{(x_1+x_2)(x_1+x_3)(x_1+x_4)} + \sqrt{(x_2+x_1)(x_2+x_3)(x_2+x_4)} + \sqrt{(x_3+x_1)(x_3+x_2)(x_3+x_4)} + \sqrt{(x_4+x_1)(x_4+x_2)(x_4+x_3)}$,\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 bound in the interval $(0, 1)$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 in high-school level. Any clues? Your helpful ideas are greatly appreciated.

We let function

$f(x_1,~x_2,~x_3,~x_4) ~=~ \sqrt{(x_1+x_2)(x_1+x_3)(x_1+x_4)} + \sqrt{(x_2+x_1)(x_2+x_3)(x_2+x_4)} + \sqrt{(x_3+x_1)(x_3+x_2)(x_3+x_4)} + \sqrt{(x_4+x_1)(x_4+x_2)(x_4+x_3)}$,

where variables $x_1,~x_2,~x_3,~x_4$ are bound in the interval $(0, 1)$ 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. Any clues? Your ideas are appreciated.

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 in high-school level. Any clues? Your helpful ideas are greatly appreciated.

Used LaTeX to rewrite formulas
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