Inequality with a+b+c=1 and $18(a^4+b^4+c^4)+6(a^2+b^2+c^2)+1\geq24(a^3+b^3+c^3)$ Let $a,b,c$ be reals with $a+b+c=1$. Show that :
$$18(a^4+b^4+c^4)+6(a^2+b^2+c^2)+1\geq24(a^3+b^3+c^3).$$
I have tried to something like this:
$$18a^4-24a^3+6a^2-12a+12\geq 0$$
$$18b^4-24b^3+6b^2-12b+12\geq 0$$
$$18c^4-24c^3+6c^2-12c+12\geq 0$$
After summing I get :
$$18(a^4+b^4+c^4)+6(a^2+b^2+c^2)+24\geq24(a^3+b^3+c^3).$$
If someone has an ideea I would gratefully appreciate.

Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative, and $abc=w^3$
Thus, since our inequality is fourth degree, we see that it's equivalent to $f(w^3)\geq0,$ 
where $f$ is a linear function.
But the linear function gets a minimal value for an extreme value of $w^3$, 
which happens for equality case of two variables.
Id est, it's enough to prove our inequality for $b=a$ and $c=1-2a,$ which gives
$$18(2a^4+(1-2a)^4)+6(2a^2+(1-2a)^2)+1\geq24(2a^3+(1-3a)^3)$$ or
$$324a^4-432a^3+180a^2-24a+1\geq0$$ or
$$(18a^2-12a+1)^2\geq0.$$
Done!
