Đáp án:
$\frac{1}{x}$ +$\frac{1}{y}$ +$\frac{1}{z}$ $\leq$ $\frac{1}{3}$
P=$\frac{x^3}{(y+z)^2}$ +$\frac{y^3}{(x+y)^2}$ +$\frac{z^3}{(x+y)^2}$
P≥$\frac{3x.y.z}{\sqrt[3]{(x+y)(y+z)(x+z)}^2}$
ta có$\sqrt[3]{(x+y)(y+z)(x+z)}$ ≤$\frac{2(x+y+z)}{3}$
=> ($\sqrt[3]{(x+y)(y+z)(x+z)}$)²≤$\frac{4(x+y+z)^2}{9}$
ta có
$\frac{x+y+z}{xyz}$ $\leq$ $\frac{1}{3}$
=>3( x+y+z)≤xyz
ta có
$\frac{1}{x}$ +$\frac{1}{y}$ +$\frac{1}{z}$ $\leq$ $\frac{1}{3}$
áp dụng BĐT cosi cho 3 số
=> $\frac{1}{x}$ +$\frac{1}{y}$ +$\frac{1}{z}$$\geq$ $\frac{3}{\sqrt[3]{xyz}}$
=> $\frac{1}{3}$ $\geq$ $\frac{3}{\sqrt[3]{xyz}}$
=> xyz≥9
=> $\frac{2(x+y+z)}{3}$≤$\frac{6(x+y+z)}{xyz}$≤2
=> $\frac{4(x+y+z)^2}{9}$≤4
=>$\frac{3x.y.z}{\sqrt[3]{(x+y)(y+z)(x+z)}^2}$≥27/4
dấu "=" xảy ra <=> x=y=z=9
