Six varibles inequality:
Given $x_1, x_2, x_3, x_4, x_5, x_6$ be non-negative real numbers satisfy \begin{cases} x_1 \ge x_2 \ge x_3 \ge x_4 \ge x_5 \ge x_6 \ge 0, \\ x_1-x_5 \le 2\sqrt{x_4\cdot x_6},\\ x_1+x_2+x_3+x_4+x_5+x_6=10.\end{cases} Prove that: $x_1^2+x_2^2+x_3^2+x_4^2+x_5^2+x_6^2 \le 22.$
Here is my attempt:I predict equality holds iff $x_1=x_2=3$ and $x_3=x_4=x_5=x_6=1.$
$\cdot$ Use AM-GM we have: $x_1 \le x_5+2\sqrt{x_4 \cdot x_6} \le x_4+x_5+x_6.$
$\Rightarrow 10=x_1+x_2+x_3+x_4+x_5+x_6 \ge x_1+\dfrac{5}{3}(x_4+x_5+x_6) \ge \dfrac{8}{3}x_1 \Rightarrow x_1 \le \dfrac{15}{4}.$
$\cdot$ Use Abel we have: \begin{align*} x_1^2+x_2^2+x_3^2+x_4^2+x_5^2+x_6^2 -22&=(x_1^2-9)+(x_2^2-9)+\sum\limits_{k=3}^6(x_k^2-1)\\&=(x_1-3)(x_1+3)+(x_2-3)(x_2+3)+\sum\limits_{k=3}^6(x_k-1)(x_k+1)\\ &=(x_1-x_2)(x_1-3)+(x_2+2-x_3)(x_1+x_2-6)+(x_3-x_4)(x_1+x_2+x_3-7)\\ &+(x_4-x_5)(x_1+x_2+x_3+x_4-8)+(x_5-x_6)(x_1+x_2+x_3+x_4+x_5-9).\end{align*}$\bullet$ Case 1: $x_6 \ge 1$, the inequality will be true. Indeed, since$$x_1+x_2-6=4-x_3-x_4-x_5-x_6 \le 0 \ \text{and} \ x_1+x_2+x_3-7=3-x_4-x_5-x_6 \le 0.$$$$x_1+x_2+x_3+x_4-8=2-x_5-x_6 \le 0 \ \text{and} \ x_1+x_2+x_3+x_4+x_5-9=1-x_6 \le 0.$$$\Rightarrow$ If $x_1 \le 3$ then inequality is obvious true. If not, $x_1 >3$ then we have:$$x_1+x_2-6=4-x_3-x_4-x_5-x_6 \le 4-\dfrac{4x_1}{3}=\dfrac{4}{3}(3-x_1) \le 0$$\begin{align*} (x_1-x_2)(x_1-3)+(x_2+2-x_3)(x_1+x_2-6) &\le (x_1-x_2)(x_1-3)+(x_2+2-x_3)\cdot \dfrac{4}{3}(3-x_1)\\ &\le \dfrac{2}{3}x_1(x_1-3)+\dfrac{8}{3}(3-x_1)\\ &=\dfrac{2}{3}(x_1-3)(x_1-4)\\ &\le 0.\end{align*}So it can imply that the inequality is true.
$\bullet$ Case 2: $0 \le x_6 <1.$ I haven't proved it yet.
I hope to see optimum solution for this case. Thanks every one.