Đặt vế trái BĐT là P
Ta có:
\(P=\sum\dfrac{1+x^2}{1+x+y^2}\ge\sum\dfrac{1+x^2}{1+\dfrac{x^2+1}{2}+y^2}=2\sum\dfrac{1+x^2}{1+x^2+2\left(1+y^2\right)}\)
Đặt \(\left(1+x^2;1+y^2;1+z^2\right)=\left(a;b;c\right)\) với \(a;b;c\ge1\)
\(\Rightarrow P\ge2\left(\dfrac{a}{a+2b}+\dfrac{b}{b+2c}+\dfrac{c}{c+2a}\right)=2\left(\dfrac{a^2}{a^2+2ab}+\dfrac{b^2}{b^2+2bc}+\dfrac{c^2}{c^2+2ca}\right)\)
\(\Rightarrow P\ge2\left(\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}\right)=2\)
Dấu "=" xảy ra khi \(a=b=c\) hay \(x=y=z=1\)
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