Ta có BĐT sau: \(\sqrt{\frac{1+a^2}{b+c}}\ge\frac{a+1}{\sqrt{2\left(b+c\right)}}\)(*)
Thật vậy, với a,b,c dương, ta có: (*)\(\Leftrightarrow\frac{1+a^2}{b+c}\ge\frac{\left(a+1\right)^2}{2\left(b+c\right)}\)
\(\Leftrightarrow\frac{1+a^2}{b+c}\ge\frac{\frac{\left(a+1\right)^2}{2}}{b+c}\Leftrightarrow1+a^2\ge\frac{a^2}{2}+a+\frac{1}{2}\)
\(\Leftrightarrow\frac{\left(a-1\right)^2}{2}\ge0\)(đúng với mọi \(a\inℝ\))
Tương tự, ta có: \(\sqrt{\frac{1+b^2}{c+a}}\ge\frac{b+1}{\sqrt{2\left(c+a\right)}}\)(2); \(\sqrt{\frac{1+c^2}{a+b}}\ge\frac{c+1}{\sqrt{2\left(a+b\right)}}\)(3)
Cộng theo vế của các BĐT (*), (2), (3), ta được:
\(\Sigma\sqrt{\frac{1+a^2}{b+c}}\ge\Sigma\frac{a+1}{\sqrt{2\left(b+c\right)}}\ge\Sigma\frac{a+1}{\frac{\left(b+c\right)+2}{2}}=\Sigma\frac{2\left(a+1\right)}{b+c+2}\)
\(=\Sigma\left(\frac{2a^2}{ab+ca+2a}+\frac{2}{b+c+2}\right)\)
\(\ge\frac{\left(a+b+c\right)^2}{\left(ab+bc+ca\right)+\left(a+b+c\right)}+\frac{9}{a+b+c+3}\)(Theo BĐT Bunhiacopxki dạng phân thức)
\(\ge\frac{\left(a+b+c\right)^2}{\frac{\left(a+b+c\right)^2}{3}+\left(a+b+c\right)}+\frac{9}{a+b+c+3}\)
\(\ge\frac{3\left(a+b+c\right)}{a+b+c+3}+\frac{9}{a+b+c+3}=\frac{3\left(a+b+c+3\right)}{a+b+c+3}=3\)
Đẳng thức xảy ra khi a = b = c = 1
