Lời giải bài 4 chuyên đề Vận dụng bất đẳng thức Côsi để tìm cực trị.
Ta có : $\frac{x^{2}}{y+z}+\frac{y+z}{4}\geq 2\sqrt{\frac{x^{2}}{y+z}.\frac{y+z}{4}}=2.\frac{x}{2}=x$
$\frac{y^{2}}{x+z}+\frac{x+z}{4}\geq 2\sqrt{\frac{y^{2}}{x+z}.\frac{x+z}{4}}=2.\frac{y}{2}=y$
$\frac{z^{2}}{x+y}+\frac{x+y}{4}\geq 2\sqrt{\frac{z^{2}}{x+y}.\frac{x+y}{4}}=2.\frac{z}{2}=z$
=> $P=\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{y+x}+\frac{y+z}{4}+\frac{x+z}{4}+\frac{x+y}{4}\geq x+y+z$
<=> $P=\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{y+x}+\frac{x+y+z}{2}\geq x+y+z$
=> $P=\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{y+x}\geq x+y+z-\frac{x+y+z}{2}\geq \frac{x+y+z}{2}=\frac{2}{2}=1$
Vậy Min P = 1 <=> $\frac{x^{2}}{y+z}=\frac{y+z}{4},\frac{y^{2}}{x+z}=\frac{x+z}{4}, \frac{z^{2}}{y+x}=\frac{y+x}{4}$
<=> $x=y=z=\frac{2}{3}$