\(\sqrt{ab\left(a+c\right)\left(b+c\right)}\le\dfrac{\left(a+b\right)}{2}.\dfrac{\left(a+c+b+c\right)}{2}=\dfrac{\left(a+b\right)\left(a+b+2c\right)}{4}=\dfrac{a^2+b^2+2ab+2ac+2bc}{4}\le\dfrac{a^2+b^2+a^2+b^2+\dfrac{4a^2+c^2}{2}+\dfrac{4b^2+c^2}{2}}{4}=\dfrac{8a^2+8b^2+2c^2}{8}=\dfrac{4a^2+4b^2+c^2}{4}\)\(\Rightarrow\dfrac{9}{\sqrt{ab\left(a+c\right)\left(b+c\right)}}-\dfrac{32}{\sqrt{4+4a^2+4b^2+c^2}}\ge\dfrac{36}{4a^2+4b^2+c^2}-\dfrac{32}{\sqrt{4+4a^2+4b^2+c^2}}\)
\(đặt:\sqrt{4+4a^2+4b^2+c^2}=t\)
\(\Rightarrow\dfrac{9}{\sqrt{ab\left(a+c\right)\left(b+c\right)}}-\dfrac{32}{\sqrt{4+4a^2+4b^2+c^2}}\ge\dfrac{36}{t^2-4}-\dfrac{32}{t}\)
\(ta\) \(đi\) \(cminh:\dfrac{36}{t^2-4}-\dfrac{32}{t}\ge-5\Leftrightarrow36t-32\left(t^2-4\right)+5\left(t^3-4t\right)\ge0\Leftrightarrow\left(t-4\right)^2\left(5t+8\right)\ge0\left(đúng\right)\)
\(\Rightarrowđpcm:\Rightarrow dấu"="\Leftrightarrow\left\{{}\begin{matrix}a=b=\dfrac{c}{2}\\t=4=\sqrt{4a^2+4b^2+c^2+4}\end{matrix}\right.\)
\(\Leftrightarrow a=b=1;c=2\)
