\(P=\sqrt{\dfrac{ab}{c+ab}}+\sqrt{\dfrac{bc}{a+bc}}+\sqrt{\dfrac{ca}{b+ca}}\)
\(P=\sqrt{\dfrac{ab}{c\left(a+b+c\right)+ab}}+\sqrt{\dfrac{bc}{a\left(a+b+c\right)+bc}}+\sqrt{\dfrac{ca}{b\left(a+b+c\right)+ca}}\)
\(P=\sqrt{\dfrac{ab}{ac+bc+c^2+ab}}+\sqrt{\dfrac{bc}{a^2+ab+ac+bc}}+\sqrt{\dfrac{ca}{ab+b^2+bc+ca}}\)
\(P=\sqrt{\dfrac{ab}{\left(a+c\right)\left(b+c\right)}}+\sqrt{\dfrac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\dfrac{ca}{\left(a+b\right)\left(b+c\right)}}\)
Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{ab}{\left(a+c\right)\left(b+c\right)}}\le\dfrac{\dfrac{a}{a+c}+\dfrac{b}{b+c}}{2}\\\sqrt{\dfrac{bc}{\left(a+b\right)\left(a+c\right)}}\le\dfrac{\dfrac{b}{a+b}+\dfrac{c}{a+c}}{2}\\\sqrt{\dfrac{ca}{\left(a+b\right)\left(b+c\right)}}\le\dfrac{\dfrac{a}{a+b}+\dfrac{c}{b+c}}{2}\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{\left(\dfrac{a}{a+c}+\dfrac{c}{a+c}\right)+\left(\dfrac{b}{b+c}+\dfrac{c}{b+c}\right)+\left(\dfrac{b}{a+b}+\dfrac{a}{a+b}\right)}{2}\)
\(\Rightarrow VT\le\dfrac{\dfrac{a+c}{a+c}+\dfrac{b+c}{b+c}+\dfrac{a+b}{a+b}}{2}=\dfrac{3}{2}\)
\(\Rightarrow P\le\dfrac{3}{2}\)
Vậy \(P_{max}=\dfrac{3}{2}\)
Dấu " = " xảy ra khi \(a=b=c=\dfrac{1}{3}\)
