$P=\frac{4 a^3+3 b^3+2 c^3-3 b^2 c}{(a+b+c)^3}\ge\frac{4}{25}$

Question

Let $a,b,c>0$. Prove that:

$P=\frac{4 a^3+3 b^3+2 c^3-3 b^2 c}{(a+b+c)^3}\ge\frac{4}{25}$

$3b^3+2c^3-3b^2c\ge b^3+c^3\ge \frac{(b+c)^3}{4}$

$\Rightarrow P\ge 4t^3+\frac{(1-t)^3}{4}$ with $t=\frac{a}{a+b+c},0<t<1$

Any other solutions?

Answer 1

After your first step it's enough to prove that: $$25(4a^3+b^3+c^3)\geq4(a+b+c)^3,$$ which is true by Holder: $$25(4a^3+b^3+c^3)=(1+2+2)^2(4a^3+b^3+c^3)\geq$$ $$\geq\left(\sqrt[3]{1^2\cdot4a^3}+\sqrt[3]{2^2\cdot b^3}+\sqrt[3]{2^2\cdot c^3}\right)^3=4(a+b+c)^3.$$