Đặt \(A=\left(\dfrac{c}{a}+\dfrac{c}{b}\right)^2\)
\(A=\left[c\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\right]^2\)
\(A=c^2\cdot\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2\)
\(A\ge\left(a^2+b^2\right)\cdot\dfrac{16}{\left(a+b\right)^2}\)
Cần cm:\(a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)(luôn đúng)
\(\Rightarrowđpcm\)