Question

\(a,b,c>0.CMR:\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ac}+\dfrac{1}{c^2+ab}\le\dfrac{a+b+c}{2abc}\)

Answer 1

\(VT=\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ac}+\dfrac{1}{c^2+ab}\le\dfrac{1}{2a\sqrt{bc}}+\dfrac{1}{2b\sqrt{ac}}+\dfrac{1}{2c\sqrt{ab}}\)

\(VT\le\dfrac{\sqrt{ab}+\sqrt{ac}+\sqrt{bc}}{2abc}\)

Mặt khác ta luôn có:

\(\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{a}-\sqrt{c}\right)^2+\left(\sqrt{b}-\sqrt{c}\right)^2\ge0\)

\(\Rightarrow2\left(a+b+c\right)-2\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)\ge0\)

\(\Rightarrow\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\le a+b+c\)

\(\Rightarrow VT\le\dfrac{a+b+c}{2abc}\)

Dấu "=" khi \(a=b=c\)