WakeUp wrote: Let $n$ be a positive integer. Prove that \[\sum_{i=1}^nx_i(1-x_i)^2\le\left(1-\frac{1}{n}\right)^2 \] for all nonnegative real numbers $x_1,x_2,\ldots ,x_n$ such that $x_1+x_2+\ldots x_n=1$. $x_i(1-x_i)^2 \le \frac{(n-1) \left(n^2 x_i-3 n x_i+2\right)}{n^3} \iff \frac{(n x_i-1)^2 (2n-nx_i-2)}{n^3} \ge 0$

Let $g(x) = x(1-x)^2$. Then $g''(x) = 6x - 4$ so $g''$ has only $1$ inflection point. So by the $n-1$ equal value principle maxima occurs when $x_1 = x_2 \dots = x_{n-1}$. So we have $(n-1)x_1 + x_n = 1 \implies x_n = 1 - (n-1)x_1$. So $\sum_{i=1}^nx_i(1-x_i)^2 \leq x_1(n-1)(1-x_1)^2 + x_n(1-x_n)^2$. It suffices to show that $x_1(n-1)(1-x_1)^2 + (1-(n-1)x_1)(n-1)^2x_1^2 \leq (1-\frac{1}{n})^2$. For convenience let $x_1 = x$. $\iff -x + nx + 3x^2 - 4nx^2 + n^2x^2 - 2nx^3 + 3n^2x^3 - n^3x^3 \leq (1-\frac{1}{n})^2$. Let $f(x) = -x + nx + 3x^2 - 4nx^2 + n^2x^2 - 2nx^3 + 3n^2x^3 - n^3x^3$. We have $f'(x) = (1-n)(3n^2x^2 + x(6 - 2n(3x+1)) - 1) = (1-n)(nx - 1)(3nx - 6x + 1) = 0 \iff x=\frac{1}{n}$ or $x=\frac{1}{6-3n}$. If $x=\frac{1}{n}$, then $f(x) = 1 - \frac{2}{n} + \frac{1}{n^2} = (1-\frac{1}{n})^2$. If $x=\frac{1}{6-3n}$, then $f(x) = \frac{-5n^2 + 14n - 9}{27(n-2)^2} \leq (1-\frac{1}{n})^2 \iff \frac{(n-1)(2n-3)}{(n-2)^2n^2} \geq 0$. $\blacksquare$

Very easy with the Lagrange method ..