Question

cho a,b,c >0 và a.b.c=1 tìm gtln của \(\frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}\)

Answer 1

\(P=\frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}\Rightarrow2P=\frac{2}{2+a}+\frac{2}{2+b}+\frac{2}{2+c}\)

\(\Rightarrow3-2P=\frac{a}{a+2}+\frac{b}{b+2}+\frac{c}{c+2}\ge\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}{a+b+c+6}\)

\(3-2P\ge\frac{a+b+c+2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)}{a+b+c+6}\ge\frac{a+b+c+6\sqrt[6]{a^2b^2c^2}}{a+b+c+6}=\frac{a+b+c+6}{a+b+c+6}=1\)

\(\Rightarrow2P\le2\Rightarrow P\le1\)