\(\Leftrightarrow\dfrac{a+b+c}{a}\left[\dfrac{\left(a+b+c\right)^2-4b\left(a+c\right)}{b\left(a+c\right)}\right]\ge\dfrac{c\left(4a+4c-5b\right)}{b\left(a+c\right)}\)
\(\Leftrightarrow\left(a+b+c\right)\left(a-b+c\right)^2\ge ac\left[4\left(a+c\right)-5b\right]\)
- Nếu \(4\left(a+c\right)\le5b\) BĐT hiển nhiên đúng
- Nếu \(4\left(a+c\right)>5b\)
Do \(ac\le\dfrac{1}{4}\left(a+c\right)^2\) nên ta chỉ cần chứng minh:
\(\left(a+b+c\right)\left(a+c-b\right)^2\ge\dfrac{1}{4}\left(a+c\right)^2\left[4\left(a+c\right)-5b\right]\)
Đặt \(\left\{{}\begin{matrix}a+c=x>0\\b=y>0\end{matrix}\right.\)
\(\Rightarrow4\left(x+y\right)\left(x-y\right)^2\ge x^2\left(4x-5y\right)\)
\(\Leftrightarrow x^2y-4xy^2+4y^3\ge0\)
\(\Leftrightarrow y\left(x-2y\right)^2\ge0\) (luôn đúng)
Dấu "=" xảy ra khi \(x=2y\) hay \(a=b=c\)
