\(a,b>0\)
\(2\left(a^2+1\right)\left(b^2+1\right)\ge\left(a+1\right)\left(b+1\right)\left(ab+1\right)\)
\(\Leftrightarrow4\left(a^2+1\right)^2\left(b^2+1\right)^2\ge\left(a+1\right)^2\left(b+1\right)^2\left(ab+1\right)^2\)
-Áp dụng BĐT \(\left(x+1\right)^2\le2\left(x^2+1\right)\) ta có:
\(\left(a+1\right)^2\left(b+1\right)^2\left(ab+1\right)^2\le2\left(a^2+1\right).2\left(b^2+1\right)\left(ab+1\right)^2=4\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)^2\)-Ta chỉ cần c/m BĐT \(4\left(a^2+1\right)^2\left(b^2+1\right)^2\ge4\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)^2\) là bài toán được c/m.
Thật vậy:
\(4\left(a^2+1\right)^2\left(b^2+1\right)^2\ge4\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)^2\)
\(\Leftrightarrow4\left(a^2+1\right)\left(b^2+1\right)\left[\left(a^2+1\right)\left(b^2+1\right)-\left(ab+1\right)^2\right]\ge0\) (*)
-Áp dụng BĐT Bunhiacopxki ta có:
\(\left(a^2+1\right)\left(b^2+1\right)\ge\left(ab+1\right)^2\)
\(\Leftrightarrow\left(a^2+1\right)\left(b^2+1\right)-\left(ab+1\right)^2\ge0\)
\(\Rightarrow\) (*) đúng do \(a^2+1;b^2+1>0\forall a,b\)
\(\Rightarrowđpcm\)
-Dấu "=" xảy ra khi \(a=b=1\)
