[Sonnhard Graubner]

Multiplying by $$abc>0$$ we have to prove that

 

$$a^3c+ab^3+bc^3\geq 3abc(a^2+b^2+c^2)$$. Since we have

$$a+b+c=1$$ we get

$$(a^3c+ab^3+bc^3)(a+b+c)\geq 3abc(a^2+b^2+c^2)$$

Using the substitution $$b=a+u,c=a+u+v$$ we get

$$4\,{a}^{3}{u}^{2}+4\,{a}^{3}uv+4\,{a}^{3}{v}^{2}+9\,{a}^{2}{u}^{3}+18
\,{a}^{2}{u}^{2}v+15\,{a}^{2}u{v}^{2}+3\,{a}^{2}{v}^{3}+7\,a{u}^{4}+20
\,a{u}^{3}v+21\,a{u}^{2}{v}^{2}+8\,au{v}^{3}+a{v}^{4}+2\,{u}^{5}+7\,{u
}^{4}v+9\,{u}^{3}{v}^{2}+5\,{u}^{2}{v}^{3}+u{v}^{4}\geq 0$$ which is true.

Sonnhard.

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