câu 9gt ⇔5x2+5(y2+z2)−9x(y+z)−18yz=0 yz≤14(y+z)2;y2+z2≥12(y+z)2 ⇒18yz−5(y2+z2)≤2(y+z)2⇒5x2−9x(y+z)≤2(y+z)2⇔(x−2(y+z))(5x+y+z)≤0\Rightarrowx≤2(y+z) P=xy2+z2−1(x+y+z)3≤2x(y+z)2−1(x+y+z)3 ≤4y+z−127(y+z)3 Đặt 1y+z=t ⇒P≤4t−127t3 Đánh giá kiểu j để P ≤16dấu '=" ⇔x=13;y=z=112
câu 9gt
⇔5x2+5(y2+z2)−9x(y+z)−18yz=0 yz≤14(y+z)2;y2+z2≥12(y+z)2 ⇒18yz−5(y2+z2)≤2(y+z)2⇒5x2−9x(y+z)≤2(y+z)2$\Leftrightarrow (x-2(y+z))(5x+y+z)\leq0
$ $\Rightarrow
x\leq 2(y+z)
P=\frac{x}{y^{2}+z^{2}} -\frac{1}{(x+y+z)^{3}} \leq \frac{2x}{(y+z)^{2}}-\frac{1}{(x+y+z)^{3}}
\leq \frac{4}{y+z}-\frac{1}{27(y+z)^{3}}
Đặt\frac{1}{y+z}=t
\Rightarrow P\leq 4t-\frac{1}{27}t^{3}
ĐánhgiákiểujđểP\leq 16
dấu′="\Leftrightarrow x=\frac{1}{3};y=z=\frac{1}{12}$