# 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? Commented 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... Commented Jul 25, 2017 at 3:02
• I posted a solution based on quadratic equations below, you can check it out :) Commented 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? Commented Jul 25, 2017 at 2:58
• @wightahtl I have fixed my post. See now please. Commented 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. Commented 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
Commented Jul 25, 2017 at 7:22