1 Ta có
$3=\frac{a}{a+x}+\frac{b}{b+y}+\frac{c}{c+z}+\frac{x}{a+x}+\frac{y}{b+y}+\frac{z}{c+z}$
$\ge 3\sqrt[3]{\frac{abc}{(a+x)(b+y)(c+z)}}+3\sqrt[3]{\frac{xyz}{(a+x)(b+y)(c+z)}}$
$\Leftrightarrow (\sqrt[3]{abc}+\sqrt[3]{xyz})^3 \le(a+x)(b+y)(c+z)$ (ok)
2 $\color{red}{Q = \frac{1}{a^4+b^2+ab^2+ab^2}+\frac{1}{b^4+a^2+a^2b+a^2b} \overset{AM-GM}\le \frac{1}{4a^2b^2}+\frac1{4a^2b^2}\\\Leftrightarrow \sqrt{2Q} \le\frac{1}{ab} \le 1 \Leftrightarrow Q \le \frac 12}$
3 $ A\ge \frac{4}{x^2+2xy+y^2}=\frac{4}{(x+y)^2} \ge 4$
4. $\color{red}{x+y \ge x^2+y^2 \ge \frac{(x+y)^2}2 \Leftrightarrow x+y \le 2}$
5. Đặt $2x=a,\sqrt 3y=b,\sqrt 3z=c$
$\Leftrightarrow 4=a^2+b^2+c^2+abc$
$\Leftrightarrow \left ( \frac{a}{2} \right )^{2}+\left(\frac{b}{2} \right )^{2}+\left ( \frac{c}{2} \right )^{2}+2.\frac{a}{2}.\frac{b}{2}.\frac{c}{2}=1$
$\Rightarrow$ tổn tại tam giác $ABC$ sao cho $\frac a2 = \cos A,\frac b2 = \cos B,\frac c2=\cos C$
Ta lại có $\cos A+\cos B+\cos C \le \frac 32 \Rightarrow$ dpcm