\(\sum\dfrac{x^2+1}{\left(z^2+1\right)+y}\ge\sum\dfrac{x^2+1}{\left(z^2+1\right)+\dfrac{y^2+1}{2}}\)
Áp dụng BĐT AM-GM ta có:
\(y\le\dfrac{y^2+1}{2}\Rightarrow\dfrac{1+x^2}{1+y+z^2}\ge\dfrac{1+x^2}{1+\dfrac{y^2+1}{2}+z^2}\)
Tương tự cho 2 BĐT còn lại thì viết lại dc thành
\(\dfrac{1+x^2}{z^2+1+\dfrac{y^2+1}{2}}+\dfrac{1+y^2}{x^2+1+\dfrac{z^2+1}{2}}+\dfrac{1+z^2}{y^2+1+\dfrac{x^2+1}{2}}\)
Đặt \(\left\{{}\begin{matrix}x^2+1=a\\y^2+1=b\\z^2+1=c\end{matrix}\right.\)\(\left(a,b,c>0\right)\) thì ta có:
\(\dfrac{a}{c+\dfrac{b}{2}}+\dfrac{b}{a+\dfrac{c}{2}}+\dfrac{c}{b+\dfrac{a}{2}}\ge2\)
\(\Leftrightarrow\dfrac{a}{2c+b}+\dfrac{b}{2a+c}+\dfrac{c}{2b+a}\ge1\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT=\dfrac{a^2}{2ac+ab}+\dfrac{b^2}{2ab+bc}+\dfrac{c^2}{2bc+ca}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{ab+bc+ca+2ab+2bc+2ca}\)
\(=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=1=VP\)