Question

Cho hai số thực a,b thuộc đoạn (0;1) và thỏa mãn \(\left(a^3+b^3\right)\left(a+b\right)-ab\left(a-1\right)\left(b-1\right)=0\) . GTLN của P=ab

Answer 1

\(\left(a^3+b^3\right)\left(a+b\right)=ab\left(a-1\right)\left(b-1\right)\)

\(\Leftrightarrow\left(\frac{a^3+b^3}{ab}\right)\left(a+b\right)=ab+1-\left(a+b\right)\)

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

Ta có: \(\left(a+b\right)\left(\frac{a^2}{b}+\frac{b^2}{a}\right)\ge\left(a+b\right)\frac{\left(a+b\right)^2}{a+b}=\left(a+b\right)^2\ge4ab\)

\(ab+1-\left(a+b\right)\le ab+1-2\sqrt{ab}\)

\(\Rightarrow4ab\le ab+1-2\sqrt{ab}\)

\(\Leftrightarrow3ab+2\sqrt{ab}-1\le0\)

\(\Leftrightarrow\left(\sqrt{ab}+1\right)\left(3\sqrt{ab}-1\right)\le0\)

\(\Leftrightarrow3\sqrt{ab}-1\le0\Leftrightarrow ab\le\frac{1}{9}\)