[tex]P\geq x^2+y^2=(x+y)^2-2xy=(x+y)^2+2(x+y)-\frac{14}{9}\\\\ +, xy\leq \frac{(x+y)^2}{4} => \frac{7}{9}\leq x+y+\frac{(x+y)^2}{4}\\\\ => (x+y)^2+4(x+y)\geq \frac{28}{9}\\\\ => (x+y+2)^2\geq \frac{64}{9}\\\\ => +, x+y+2\geq \frac{8}{3} => x+y\geq \frac{2}{3}\\\\ +, x+y+2\leq \frac{-8}{3} => x+y\leq \frac{-14}{3}\\\\ *, x+y\geq \frac{2}{3} \\\\ => P\geq (x+y)^2+4(x+y)-\frac{14}{9}\geq \frac{4}{9}+\frac{24}{9}-\frac{14}{9}=\frac{14}{9} (1)\\\\ *, x+y\leq \frac{-14}{3}<0\\\\ => P\geq (x+y+2)^2-\frac{32}{9}\\\\ +, x+y+2\leq \frac{-8}{3}<0 => (x+y+2)^2\geq \frac{64}{9}\\\\ => P\geq \frac{32}{9} (2)\\\\ (1);(2)=> P\geq \frac{14}{9}[/tex]
dấu "=" <=> x=y=1/3 và z=0
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
