\(VT=\sqrt{\dfrac{b^2c^2}{a\left(a+b+c\right)+bc}}+\sqrt{\dfrac{a^2c^2}{b\left(a+b+c\right)+ac}}+\sqrt{\dfrac{a^2b^2}{c\left(a+b+c\right)+ab}}\)
\(VT=\sqrt{\dfrac{b^2c^2}{a^2+ab+ac+bc}}+\sqrt{\dfrac{a^2c^2}{ab+b^2+bc+ca}}+\sqrt{\dfrac{a^2b^2}{ca+bc+c^2+ab}}\)
\(VT=\sqrt{\dfrac{b^2c^2}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\dfrac{a^2c^2}{\left(b+c\right)\left(a+b\right)}}+\sqrt{\dfrac{a^2b^2}{\left(c+a\right)\left(c+b\right)}}\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{b^2c^2}{\left(a+b\right)\left(a+c\right)}}\le\dfrac{\dfrac{bc}{a+b}+\dfrac{bc}{a+c}}{2}\\\sqrt{\dfrac{a^2c^2}{\left(a+b\right)\left(b+c\right)}}\le\dfrac{\dfrac{ca}{a+b}+\dfrac{ca}{b+c}}{2}\\\sqrt{\dfrac{a^2b^2}{\left(c+a\right)\left(c+b\right)}}\le\dfrac{\dfrac{ab}{c+a}+\dfrac{ab}{c+b}}{2}\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{\left(\dfrac{bc}{a+b}+\dfrac{ca}{a+b}\right)+\left(\dfrac{ca}{b+c}+\dfrac{ab}{b+c}\right)+\left(\dfrac{bc}{c+a}+\dfrac{ab}{c+a}\right)}{2}\)
\(\Rightarrow VT\le\dfrac{\left[\dfrac{c\left(a+b\right)}{a+b}\right]+\left[\dfrac{a\left(b+c\right)}{b+c}\right]+\left[\dfrac{b\left(c+a\right)}{c+a}\right]}{2}\)
\(\Rightarrow VT\le\dfrac{a+b+c}{2}=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{bc}{\sqrt{a+bc}}+\dfrac{ac}{\sqrt{b+ca}}+\dfrac{ab}{\sqrt{c+ab}}\le\dfrac{1}{2}\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c=\dfrac{1}{3}\)
