Đặt vế trái là P
\(P=\left(\frac{1}{1+\frac{b}{a}}\right)^2+\left(\frac{1}{1+\frac{c}{b}}\right)^2+4\left(\frac{1}{1+\frac{a}{c}}\right)^2\)
Đặt \(\left\{{}\begin{matrix}\frac{b}{a}=x>0\\\frac{c}{b}=y>0\end{matrix}\right.\) \(\Rightarrow xy=\frac{c}{a}\ge1\)
\(P=\frac{1}{\left(1+x\right)^2}+\frac{1}{\left(1+y\right)^2}+4\left(\frac{1}{1+\frac{1}{xy}}\right)^2=\frac{1}{\left(1+x\right)^2}+\frac{1}{\left(1+y\right)^2}+4\left(\frac{xy}{1+xy}\right)^2\)
\(P\ge\frac{1}{1+xy}+4\left(\frac{xy}{1+xy}\right)^2\)
Đặt \(xy=t\ge1\Rightarrow P\ge\frac{1}{1+t}+4\left(\frac{t}{1+t}\right)^2\)
Ta chỉ cần chứng minh \(\frac{1}{1+t}+4\left(\frac{t}{1+t}\right)^2\ge\frac{3}{2}\)
\(\Leftrightarrow1+t+4t^2\ge\frac{3}{2}\left(1+t\right)^2\)
\(\Leftrightarrow8t^2+2t+2\ge3t^2+6t+3\)
\(\Leftrightarrow5t^2-4t-1\ge0\Leftrightarrow\left(t-1\right)\left(5t+1\right)\ge0\) (luôn đúng \(\forall t\ge1\))
Dấu "=" xảy ra khi \(t=1\) hay \(a=b=c\)