Áp dụng BĐT Cauchy
\(\Rightarrow\left(a+b+c\right)\left(ab+bc+ac\right)\ge9abc\)
\(\Rightarrow\sqrt{\dfrac{\left(a+b+c\right)\left(ab+bc+ac\right)}{abc}}\ge3\)
\(\Rightarrow P\ge3+\dfrac{4bc}{\left(b+c\right)^2}\)
Ta cần tìm Min của \(3+\dfrac{4bc}{\left(b+c\right)^2}\)
Không mất tính tổng quát giả sử \(b\ge c\)
\(\Rightarrow b+c\le2b\)\(\Leftrightarrow\left(b+c\right)^2\le4b^2\Leftrightarrow\dfrac{4bc}{\left(b+c\right)^2}\ge\dfrac{c}{b}\)
\(b\ge c\Rightarrow\dfrac{c}{b}\ge1\)
Vậy \(3+\dfrac{4bc}{\left(b+c\right)^2}\ge4\)
Dấu đẳng thức xảy ra khi a = b = c
Áp dụng BĐT bunyakovsky và AM -GM ta có:
\(\sqrt{\dfrac{\left[a+\left(b+c\right)\right]\left[bc+a\left(b+c\right)\right]}{abc}}\ge\sqrt{\dfrac{a\left(\sqrt{bc}+b+c\right)^2}{abc}}=\dfrac{\sqrt{bc}+b+c}{\sqrt{bc}}=1+\dfrac{b+c}{\sqrt{bc}}\)
\(LHS\ge1+\dfrac{b+c}{2\sqrt{bc}}+\dfrac{b+c}{2\sqrt{bc}}+\dfrac{4bc}{\left(b+c\right)^2}\ge1+3\sqrt[3]{\dfrac{4bc\left(b+c\right)^2}{4bc\left(b+c\right)^2}}=4\)
Dấu = xảy ra khi a=b=c
