đề thế này à .Cho x,y,z>0 thỏa mãn x+y+z=xyz.
Chứng minh \(\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}+\frac{1}{\sqrt{1+z^2}}\le\frac{3}{2}\)
\(Gt\Rightarrow\text{ ∑}\frac{1}{xy}=1\).Đặt \(\frac{1}{x}=a;\frac{1}{y}=b;\frac{1}{z}=c\) ta có:
\(ab+bc+ca=1\)
\(\text{∑}\frac{1}{\sqrt{1+x^2}}=\text{∑}\frac{a}{\sqrt{1+a^2}}=\text{∑}\frac{a}{\sqrt{ab+bc+ca+a^2}}\)
\(=\text{∑}\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\frac{1}{2}\text{∑}\left(\frac{a}{a+b}+\frac{a}{c+a}\right)=\frac{3}{2}\)