Question

Cho các số thực dương a, b. Tìm GTLN của biểu thức

P = (a + b)(\(\frac{1}{a^3+b}+\frac{1}{b^3+a}\))\(-\frac{1}{ab}\)

Answer 1

\(\left(a^3+b\right)\left(\frac{1}{a}+b\right)\ge\left(a+b\right)^2\Rightarrow\frac{1}{a^3+b}\le\frac{ab+1}{a\left(a+b\right)^2}\)

Tương tự: \(\frac{1}{b^3+a}\le\frac{ab+1}{b\left(a+b\right)^2}\)

\(\Rightarrow P\le\left(a+b\right)\left(\frac{ab+1}{a\left(a+b\right)^2}+\frac{ab+1}{b\left(a+b\right)^2}\right)-\frac{1}{ab}\)

\(P\le\frac{ab+1}{a\left(a+b\right)}+\frac{ab+1}{b\left(a+b\right)}-\frac{1}{ab}=\left(\frac{ab+1}{a+b}\right)\left(\frac{1}{a}+\frac{1}{b}\right)-\frac{1}{ab}\)

\(P\le\frac{\left(ab+1\right)\left(a+b\right)}{ab\left(a+b\right)}-\frac{1}{ab}=\frac{ab+1}{ab}-\frac{1}{ab}=1\)

\(P_{max}=1\) khi \(a=b=1\)