Question

Cho a,b thỏa mãn \(a+b\ne0\)

Chứng minh rằng\(a^2+b^2+\left(\frac{ab+1}{a+b}\right)^2\ge2\)

Answer 1

BĐT tương đương

\(a^2+b^2+\frac{a^2b^2+2ab+1}{\left(a+b\right)^2}\ge2\)

<=>\(\left(a+b\right)^2-2+\frac{1}{\left(a+b\right)^2}+\frac{a^2b^2}{\left(a+b\right)^2}+\frac{2ab}{\left(a+b\right)^2}-2ab\ge0\)

<=>\(\left(a+b\right)^2-2.\left(a+b\right).\frac{1}{a+b}+\frac{a^2b^2}{\left(a+b\right)^2}-2.\left(ab-\frac{ab}{\left(a+b\right)^2}\right)\ge0\)

<=>\(\left(a+b-\frac{1}{a+b}\right)^2+\frac{a^2b^2}{\left(a+b\right)^2}-2.\left(\frac{ab\left(a+b\right)^2-ab}{\left(a+b\right)^2}\right)\ge0\)

<=>\(\left(\frac{\left(a+b\right)^2-1}{a+b}\right)^2+\frac{a^2b^2}{\left(a+b\right)^2}-2.\left(\frac{ab\left[\left(a+b\right)^2-1\right]}{\left(a+b\right)\left(a+b\right)}\right)\ge0\)

<=>\(\left(\frac{\left(a+b\right)^2-1}{a+b}\right)^2+\frac{a^2b^2}{\left(a+b\right)^2}-2.\frac{\left(a+b\right)^2-1}{a+b}.\frac{ab}{a+b}\ge0\)

<=>\(\left(\frac{\left(a+b\right)^2-1}{a+b}-\frac{ab}{a+b}\right)^2\ge0\left(\text{luôn đúng}\right)\)

=> dpcm