Đang suy nghĩ
\(P=\frac{ab}{\sqrt{c+ab}}+\frac{bc}{\sqrt{a+bc}}+\frac{ca}{\sqrt{b+ca}}\)
\(=\sqrt{\frac{a^2b^2}{c\left(a+b+c\right)+ab}}+\sqrt{\frac{b^2c^{^2}}{a\left(a+b+c\right)+bc}}+\sqrt{\frac{c^2a^2}{b\left(a+b+c\right)+ca}}\)
\(=\sqrt{\frac{a^2b^2}{\left(c+a\right)\left(c+b\right)}}+\sqrt{\frac{b^2c^2}{\left(a+b\right)\left(c+a\right)}}+\sqrt{\frac{c^2a^2}{\left(a+b\right)\left(b+c\right)}}\)
Áp dụng BĐT cô - si ta có :
\(\sum\sqrt{\frac{a^2b^2}{\left(c+a\right)\left(c+b\right)}}\le\sum\frac{\frac{ab}{c+a}+\frac{ab}{c+b}}{2}=\sum\frac{b\left(a+c\right)}{\frac{a+c}{2}}=\sum\frac{b}{2}=\frac{a+b+c}{2}=\frac{1}{2}\)
Vậy \(MAX_P=\frac{1}{2}\) khi \(a=b=c=\frac{1}{3}\)
