Cho các số thực x,y thỏa mãn $x^{2016}+y^{2016}\leq2$. Tìm GTNN của biểu thức: $P=x(y+1)^{2}+y(x+1)^{2}$
Dùng bdt cosi cho 2016 số dương, ta có
$\left.\begin{matrix} x^{2016}+1+1+...+1 \ge 2016|x|\\ y^{2016}+1+1+...+1 \ge 2016|y|\end{matrix}\right\}\Rightarrow x^{2016}+y^{2016}+4030 \ge 2016(|x|+|y|)$
Mà $x^{2016}+y^{2016} \le 2$
Suy ra $|x|+|y| \le 2$
Ta có $P=xy(x+y)+4xy+(x+y)$
$=(xy+1)(x+y+4)-4$
Vì $|x|+|y| \le 2\Rightarrow |x| \le 2,|y| \le 2\Rightarrow x+y \ge -4$
Do đó $x+y+4 \ge0$

Lại có $2 \ge |x|+|y| \ge 2\sqrt{|xy|}\Rightarrow |xy| \le 1$
Do đó $xy+1 \ge 0$
Từ 2 điều trên suy ra $P \ge -4$
Mặt khác khi thay $x=1,y=-1$ (thỏa đk) vào $P$ thì $P=-4$
Vậy ta kết luật $P=-4$ là GTNN, đạt được khi $x=1,y=-1$

ở đâu nhỉ –  tran85295 17-08-16 07:39 PM
Dòng thứ 8 là sao nhỉ –  pinocchio 17-08-16 05:51 PM

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