# Find the maximum of $x^2_{1}+x^2_{2}+\cdots+x^2_{n}+\sqrt{x_{1}x_{2}\cdots x_{n}}$ if $x_i\ge0$ and $x_{1}+x_{2}+\cdots+x_{n}=1$

Give the postive integer $n\ge 2$,and $x_{i}\ge 0$,such $$x_{1}+x_{2}+\cdots+x_{n}=1$$ Find the maximum of the value $$x^2_{1}+x^2_{2}+\cdots+x^2_{n}+\sqrt{x_{1}x_{2}\cdots x_{n}}$$

I try $$x_{1}x_{2}\cdots x_{n}\le\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)^n=\dfrac{1}{n^n}$$

• What about AM-GM meets Cauchy-Schwarz? Jul 25, 2017 at 2:25
• You can do it with LM. I worked it out and see that it is possible, and the issue is at the boundary which can be taken care with limit... Jul 25, 2017 at 3:02
• I posted a solution based on quadratic equations below, you can check it out :) Jul 25, 2017 at 5:28

Also we can make the following.

Let $x_1=x_2=...=x_{n-1}=0$ and $x_n=1$.

Hence, we get a value $1$ and it's a maximal value because for this we need to probe that $$\sum_{i=1}^nx_i^2+\sqrt{\prod_{i=1}^nx_i}\leq\left(\sum_{i=1}^nx_i\right)^2,$$ which is true by AM-GM.

Indeed, let $\prod\limits_{i=1}^nx_i=w^n$, where $w\geq0$, $\sum\limits_{1\leq i<j\leq n}x_ix_j=\frac{n(n-1)}{2}v^2$, where $v\geq0$, and $\sum\limits_{i=1}^nx_1=nu$.

Hence, by AM-GM $u\geq w$,$v\geq w$ and we need to prove that $$n^2(n-1)^2v^4\geq w^n$$ or $$n^2(n-1)^2v^4\cdot n^{n-4}u^{n-4}\geq w^n,$$ which is obviously true for $n\geq4$.

Thus, it's remains to understand, what happens for $n=2$ and for $n=3$, which is for you.

• Hello,This is contest problem,have without EV methods? Jul 25, 2017 at 2:58
• @wightahtl I have fixed my post. See now please. Jul 25, 2017 at 3:34

Let $x_1\geq x_2 \geq ... \geq x_n$ without loss of generality. Notice that $$\begin{split}F(x_1,...)=\left(\sum_{i=0}^n x_i\right)^2-\sum_{i=0}^n x_i^2 - \sqrt{\prod_{i=0}^n x_i} &= 2\cdot\sum_{i<j}x_ix_j-\sqrt{\prod_{i=0}^n x_i} \\ &=\left(2\sum_{i>1}x_i\right)\cdot x_1-\sqrt{\prod_{i=2}^n x_i}\cdot\sqrt{x_1} + 2\sum_{i>j>1}x_ix_j\end{split}$$

This is a quadratic function of $\sqrt{x_1}$. We can then show that $F\geq0$ by showing that the quadratic function $$f(y) = \left(2\sum_{i>1}x_i\right)\cdot y^2-\sqrt{\prod_{i=2}^n x_i}\cdot y + 2\sum_{i>j>1}x_ix_j$$ satisfies $\triangle \leq 0$ for any $0\leq\sum_{i=2}^n x_i \leq 1$. This can be shown by $$\triangle = \prod_{i=2}^n x_i-16\sum_{i>1}x_i\cdot\sum_{i>j>1}x_ix_j\leq \prod_{i=2}^nx_i-16\cdot x_2^2\cdot \sum_{j>2}x_j \leq x_2\cdot\left(x_3x_4...x_n-16\cdot x_2x_3\right)\leq 0$$ Where the last inequality is because $16x_2\geq x_2\geq x_4\geq x_4x_5...$ Hence, this implies that $$F(x_1,...)\geq 0\implies x_1^2+...+x_n^2+\sqrt{x_1...x_n}\leq\left(x_1+...+x_n\right)^2 = 1$$ which occurs when $(x_1,...,x_n) = (1,0,...0)$ or some permutation of this.

When the function is symmetric in several variables, I believe it can be shown the extrema occur at points where the $x_i$ are all equal. Then $x_i=\frac1n$ for each i.Then there's the matter of whether it's a max or a min. Putting that aside: Well, we get $$n×\frac{1}{n^2}+\sqrt {\left(\frac1n \right)^n}$$ $$=\frac1n + \sqrt {\frac {1}{n^n}}$$. If there is indeed a max, this should be it. For n=2, we get 1. For n=3 we get $\frac13 + \sqrt {\frac1{27}}$...But this must be a min; since (0,0,1) gives 1.

• To be honest, this answer does not warrant an upvote, and the answers that deserve an upvote are the ones below. Why? your post does not prove any thing. Jul 25, 2017 at 6:53
• And it starts by the assertion "When the function is symmetric in several variables, ... the extrema occur at points where the xi are all equal", squarely wrong.
– Did
Jul 25, 2017 at 7:22