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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.

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    $\begingroup$ Homogenize it. $6(a^2+b^2+c^2) = 6(a^2+b^2+c^2)(a+b+c)^2, a^3+b^3+c^3= (a^3+b^3+c^3)(a+b+c)$, and look out for Muirhead and Schur inequalities forms. $\endgroup$ – DeepSea Feb 21 at 19:22
  • $\begingroup$ Your two equations are the same, but somehow you put "...+24" instead of "...+1". $\endgroup$ – Infiaria Feb 21 at 19:24
  • $\begingroup$ And $1=(a+b+c)^4$? $\endgroup$ – mathlearning Feb 21 at 19:26
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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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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!

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