<s>[INST] 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.
 [/INST] Writing your inequality in the form $$\frac{18(a^4+b^4+c^4)}{a+b+c}+6(a^2+b^2+c^2)(a+b+c)+(a+b+c)^3-24(a^3+b^3+c^3)\geq 0$$ so we get that this is equivalent to
$$\frac{\left(a^2-4 a b-4 a c+b^2-4 b
   c+c^2\right)^2}{a+b+c}\geq 0$$ which is true.
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