Question

[Hoàn Minh]

Cho \(a,b,c>0\). Tìm min:

\(P=\dfrac{a^2}{b^2+c^2}+\dfrac{b^2}{c^2+a^2}+\dfrac{c^2}{a^2+b^2}-\dfrac{12abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\) 

Answer 1

[Nguyễn Việt Lâm]

\(P=\dfrac{a^4}{a^2b^2+a^2c^4}+\dfrac{b^4}{b^2c^2+a^2b^2}+\dfrac{c^4}{a^2+b^2}-\dfrac{12abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

\(P\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(a^2b^2+b^2c^2+c^2a^2\right)}-\dfrac{12abc}{2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ac}}\)

\(P\ge\dfrac{3\left(a^2b^2+b^2c^2+c^2a^2\right)}{2\left(a^2b^2+b^2c^2+c^2a^2\right)}-\dfrac{3}{2}=0\)

\(P_{min}=0\) khi \(a=b=c\)