Inequality with a+b+c=1 and $18(a^4+b^4+c^4)+6(a^2+b^2+c^2)+1\geq24(a^3+b^3+c^3)$

Question

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.

Answer 1

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.

Answer 2

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!

Answer 3

Alternative proof:

We use the pqr method.

Let $p = a + b + c = 1, q = ab + bc + ca, r = abc$.

It is easy to obtain $$a^2 + b^2 + c^2 = p^2 -2 q,$$ and $$a^3 + b^3 + c^3 = p(p^2 - 3q) + 3r,$$ and $$a^4 + b^4 + c^4 = (p^2 - 2q)^2 - 2(q^2 - 2pr).$$

$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$, and $(a^2 + b^2 + c^2)^2 = a^4 + b^4 + c^4 + 2(a^2b^2 + b^2c^2 + c^2a^2)$ $= a^4 + b^4 + c^4 + 2[(ab + bc + ca)^2 - 2abc(a + b + c)]$.

The desired inequality is written as $$18\cdot [(1 - 2q)^2 - 2(q^2 - 2r)] + 6(1 - 2q) + 1 \ge 24\cdot (1 - 3q + 3r) $$ or $$36q^2 - 12q + 1 \ge 0$$ or $$(6q - 1)^2 \ge 0$$ which is true.

We are done.