Considering the right expression we have \(\dfrac{a^2bc}{2a+b+c}+\dfrac{ab^2c}{a+2b+c}+\dfrac{abc^2}{a+b+2c}\)
\(=abc\left(\dfrac{a}{a+b+a+c}+\dfrac{b}{a+b+b+c}+\dfrac{c}{a+c+b+c}\right)\)
Apply Cauchy Schwarz inequality we have the formula \(\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge\dfrac{1}{a+b}\)
Apply this formula to the following expression we have
\(\dfrac{a}{a+b+a+c}\le\dfrac{a}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\)\(;\dfrac{b}{a+b+b+c}\le\dfrac{b}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)\)
\(\dfrac{c}{a+c+c+a}\le\dfrac{c}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{c+a}\right)\)
\(\Rightarrow\dfrac{a}{a+b+a+c}+\dfrac{b}{a+b+b+c}+\dfrac{c}{a+c+c+a}\le\dfrac{a+b}{4\left(a+b\right)}+\dfrac{b+c}{4\left(b+c\right)}+\dfrac{c+a}{4\left(c+a\right)}=\dfrac{3}{4}\)
\(\Rightarrow abc\left(\dfrac{a}{a+b+a+c}+\dfrac{b}{a+b+b+c}+\dfrac{c}{a+c+b+c}\right)\le\dfrac{3abc}{4}\)
We now need to prove that \(\dfrac{a^3b}{3a+b}+\dfrac{b^3c}{3b+c}+\dfrac{c^3a}{3c+a}\ge\dfrac{3abc}{4}\)
\(\Leftrightarrow\dfrac{a^3b}{3a+b}-\dfrac{abc}{4}+\dfrac{b^3c}{3b+c}-\dfrac{abc}{4}+\dfrac{c^3a}{3c+a}-\dfrac{abc}{4}\ge0\)
\(\Leftrightarrow\dfrac{a^3bc}{c\left(3a+b\right)}-\dfrac{abc}{4}+\dfrac{ab^3c}{a\left(3b+c\right)}-\dfrac{abc}{4}+\dfrac{c^3ab}{b\left(3c+a\right)}-\dfrac{abc}{4}\ge0\)
\(\Leftrightarrow abc\left[\dfrac{a^2}{c\left(3a+b\right)}-\dfrac{1}{4}\right]+abc\left[\dfrac{b^2}{a\left(3b+c\right)}-\dfrac{1}{4}\right]+abc\left[\dfrac{c^2}{b\left(3c+a\right)}-\dfrac{1}{4}\right]\ge0\)
\(\Leftrightarrow abc\left[\dfrac{a^2}{c\left(3a+b\right)}+\dfrac{b^2}{a\left(3b+c\right)}+\dfrac{c^2}{b\left(3c+a\right)}-\dfrac{3}{4}\right]\ge0\)
\(\Leftrightarrow\dfrac{a^2}{c\left(3a+b\right)}+\dfrac{b^2}{a\left(3b+c\right)}+\dfrac{c^2}{b\left(3c+a\right)}\ge\dfrac{3}{4}\)
Apply Cauchy Schwarz inequality Engel form we have
\(\dfrac{a^2}{c\left(3a+b\right)}+\dfrac{b^2}{a\left(3b+c\right)}+\dfrac{c^2}{b\left(3c+a\right)}\ge\dfrac{\left(a+b+c\right)^2}{4\left(ab+bc+ac\right)}\)
According to the Cauchy's consequence, we have \(\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\)
\(\Rightarrow\dfrac{\left(a+b+c\right)^2}{4\left(ab+bc+ac\right)}\ge\dfrac{3\left(ab+bc+ac\right)}{4\left(ab+bc+ac\right)}=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{a^2}{c\left(3a+b\right)}+\dfrac{b^2}{a\left(3b+c\right)}+\dfrac{c^2}{b\left(3c+a\right)}\ge\dfrac{3}{4}\)
So now we have \(\dfrac{a^3b}{3a+b}+\dfrac{b^3c}{3b+c}+\dfrac{c^3a}{3c+a}\ge\dfrac{3abc}{4}\)
But \(abc\left(\dfrac{a}{a+b+a+c}+\dfrac{b}{a+b+b+c}+\dfrac{c}{a+c+b+c}\right)\le\dfrac{3abc}{4}\)
\(\Rightarrow\dfrac{a^3b}{3a+b}+\dfrac{b^3c}{3b+c}+\dfrac{c^3a}{3c+a}\ge\dfrac{a^2bc}{2a+b+c}+\dfrac{ab^2c}{a+2b+c}+\dfrac{abc^2}{a+b+2c}\) ( things must be proven.)
