Question

Cho a,b,c>0 . CM :

\(\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ca}+\dfrac{1}{c^2+ab}\le\dfrac{a+b+c}{2abc}\)

Answer 1

Aps dụng bđt Cauchy :

\(\dfrac{1}{2}\left(\dfrac{1}{ab}+\dfrac{1}{ac}\right)\ge\dfrac{1}{a\sqrt{bc}}\ge\dfrac{2}{a^2+bc}\\ \dfrac{1}{2}\left(\dfrac{1}{bc}+\dfrac{1}{ba}\right)\ge\dfrac{1}{b\sqrt{ac}}\ge\dfrac{2}{b^2+ac}\\ \dfrac{1}{2}\left(\dfrac{1}{ac}+\dfrac{1}{cb}\right)\ge\dfrac{1}{c\sqrt{ab}}\ge\dfrac{2}{c^2+ab}\)

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\(\Rightarrow\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ac}+\dfrac{1}{c^2+ab}\le\dfrac{1}{2}\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=\dfrac{a+b+c}{2abc}\)