Ta có: \(a^2+2b^2+3=a^2+b^2+b^2+1+2\)
Áp dụng BĐT AM-GM, ta có:
\(a^2+b^2\ge2\sqrt{a^2b^2}=2ab\)
\(b^2+1\ge2\sqrt{b^21}=2b\)
\(\Rightarrow a^2+b^2+b^2+1+2\ge2ab+2b+2\)
\(\Rightarrow\dfrac{1}{a^2+2b^2+3}\le\dfrac{1}{2ab+2b+2}\)
Chứng minh tương tự ta được: \(\dfrac{1}{b^2+2c^2+3}\le\dfrac{1}{2bc+2c+2}\); \(\dfrac{1}{c^2+2a^2+3}1\le\dfrac{1}{2ac+2a+2}\)
\(\Rightarrow\dfrac{1}{a^2+2b^2+3}+\dfrac{1}{b^2+2c^2+3}+\dfrac{1}{c^2+2a^2+3}\le\dfrac{1}{2ab+2b+2}+\dfrac{1}{2bc+2c+2}+\dfrac{1}{2ac+2a+2}=\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{1}{bc+c+1}+\dfrac{1}{ac+a+1}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{abc}{bc+c+abc}+\dfrac{abc}{ac+a.abc+abc}\right)=\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{ab}{b+1+ab}+\dfrac{b}{1+ab+b}\right)=\dfrac{1}{2}.\dfrac{1+ab+b}{ab+b+1}=\dfrac{1}{2}\)
\(\RightarrowĐPCM\)
