Chứng minh:
\(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\) (1) với a; b \(\ge\)1
Thật vậy:
(1) <=> \(\frac{2+a^2+b^2}{1+a^2+b^2+a^2b^2}\ge\frac{2}{1+ab}\)
<=> \(2+a^2+b^2+2ab+a^3b+ab^3\ge2+2a^2+2b^2+2a^2b^2\)
<=> \(a^3b+ab^3+2ab-a^2-b^2-2a^2b^2\ge0\)
<=> \(ab\left(a^2+b^2-2ab\right)-\left(a^2+b^2-2ab\right)\ge0\)
<=> \(\left(ab-1\right)\left(a-b\right)^2\ge0\)đúng với a; b \(\ge\)1
Vậy (1) đúng
Áp dụng ta có:
\(\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}+\frac{1}{1+abc}\ge\frac{2}{1+ab}+\frac{2}{1+c\sqrt{abc}}\)
\(=2\left(\frac{1}{1+\left(\sqrt{ab}\right)^2}+\frac{1}{1+\left(\sqrt{c\sqrt{abc}}\right)^2}\right)\ge2.\frac{2}{1+\sqrt{ab}.\sqrt{c\sqrt{abc}}}=\frac{4}{1+\sqrt{abc\sqrt{abc}}}\)
\(\ge\frac{4}{1+\sqrt{abc.abc}}=\frac{4}{1+abc}\)
=> \(\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}\ge\frac{3}{1+abc}\)
Dấu "=" xảy ra <=> a = b = c
