Đề:Cho a,b,c>0 thỏa mãn abc=1.Tìm Max:$P=\frac{1}{\sqrt{a^5-a^2+3ab+6}}+\frac{1}{\sqrt{b^5-b^2+3bc+6}}+\frac{1}{\sqrt{c^5-c^2+3ac+6}}$
$P^2\leq 3(\sum \frac{1}{a^5-a^2+3ab+6}) $
Ta có các đánh giá sau:
$a^5+a+1\geq 3a^2,a^2+1\geq 2a\Rightarrow a^5-a^2+1\geq 4a-2-a=3a-2$
$\Leftrightarrow a^5-a^2+3ab+6\geq 3(a+ab+1)$
$\Rightarrow P^2\leq \sum \frac{1}{a+ab+1}$
Ta lại có $\frac{1}{a+ab+1}+\frac{1}{b+bc+1}+\frac{1}{c+ac+1}=\frac{bc+1}{1+bc+b}+\frac{1}{c+ac+1}=\frac{c+ac}{1+c+ac}+\frac{1}{c+ac+1}=1$(Chú ý giả thiết abc=1)
Vậy Max=1 khi a=b=c=1