Question

Cho \(a,b,c\in R^+,\)CMR:

\(\dfrac{2\left(a^3+b^3+c^3\right)}{abc}+\dfrac{9\left(a+b+c\right)^2}{a^2+b^2+c^2}\ge33\)

Answer 1

\(BDT\Leftrightarrow2\left[\dfrac{a^3+b^3+c^2}{abc}-3\right]+9\left[\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2}-3\right]\ge0\)

\(\Leftrightarrow\dfrac{\left(a+b+c\right)\sum\left(a-b\right)^2}{abc}+\dfrac{-9\sum\left(a-b\right)^2}{a^2+b^2+c^2}\ge0\)

\(\Leftrightarrow\sum\left(a-b\right)^2\left(\dfrac{a+b+c}{abc}-\dfrac{9}{a^2+b^2+c^2}\right)\ge0\)

\(\Leftrightarrow\sum\left(a-b\right)^2.\dfrac{\sum\left(a-b\right)^2.\left(a+b+3c\right)}{2abc\left(a^2+b^2+c^2\right)}\ge0\) (đúng)