# A four-variable maximization problem [closed]

We let function

\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}

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.

• MSE is a right forum for such type questions. – user64494 Nov 18 at 15:26
• @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"? – Yemon Choi Nov 19 at 10:06

## 2 Answers

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{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}

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{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}

with equality attained only when $$x_1=x_2$$.

Adding the two inequalities we obtain

\begin{aligned} f(x_1,~x_2,~x_3,~x_4) ~&\leq \sqrt{(x_1+x_2)}+\sqrt{(x_3+x_4)} \end{aligned}

By Jensen's Inequality since $$-\sqrt{x}$$ is convex for $$0\leq x \leq 1$$ we have

\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}

with equality attained only when $$(x_1+x_2)=(x_3+x_4)=\frac{1}{2}$$.

Hence

\begin{aligned} f(x_1,~x_2,~x_3,~x_4) ~&\leq \sqrt{2} \end{aligned}

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 $$x_1=x_2=x_3=x_4=\frac{1}{4}.$$

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 $$a=b=c=d=\frac14$$.