Do cả 2 vế cùng dương,ta bình phương 2 vế:
\(bđt\Leftrightarrow a^2+1+b^2+1+c^2+1+2\sqrt{\left(1+a^2\right)\left(1+b^2\right)}+2\sqrt{\left(1+b^2\right)\left(1+c^2\right)}+2\sqrt{\left(1+c^2\right)\left(1+a^2\right)}\ge2\left(a+b+c\right)+2\sqrt{\left(a+b\right)\left(b+c\right)}+2\sqrt{\left(b+c\right)\left(c+a\right)}+2\sqrt{\left(c+a\right)\left(a+b\right)}\)Ta chứng minh từng bđt:
\(a^2+1+b^2+1+c^2+1\ge2\left(a+b+c\right)\Rightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\)(1)
Cần cm: \(\sqrt{\left(1+a^2\right)\left(1+b^2\right)}+\sqrt{\left(1+b^2\right)\left(1+c^2\right)}+\sqrt{\left(1+c^2\right)\left(1+a^2\right)}\ge\sqrt{\left(a+b\right)\left(b+c\right)}+\sqrt{\left(b+c\right)\left(c+a\right)}+\sqrt{\left(c+a\right)\left(a+b\right)}\)
Thật vậy theo Bunyakovsky: \(\sqrt{\left(1+a^2\right)\left(b^2+1\right)}+\sqrt{\left(1+b^2\right)\left(c^2+1\right)}+\sqrt{\left(1+c^2\right)\left(a^2+1\right)}\ge a+b+b+c+c+a=2\left(a+b+c\right)\)
Theo AM-GM: \(\sqrt{\left(a+b\right)\left(b+c\right)}+\sqrt{\left(b+c\right)\left(a+c\right)}+\sqrt{\left(a+c\right)\left(a+b\right)}\le\frac{a+b+b+c+b+c+a+c+a+c+a+b}{2}=2\left(a+b+c\right)\) (2)
TTừ 1;2 ta có bđt được cm
