Cô-si : \(\sqrt{\left(a+b\right)\left(b+c\right)}\le\frac{a+b+b+c}{2}=\frac{a+2b+c}{2}\)
Ta sẽ chứng minh \(VT\le6=\Sigma_{cyc}\frac{a+2b+c}{2}\) . Ta có:
\(VP-VT=\Sigma_{cyc}\frac{\left(a-b\right)^2}{2\left(\sqrt{c+a}+\sqrt{b+c}\right)^2}\ge0\)
Từ đó..