<s>[INST] Inequality $(a+b+c)(ab+bc+ca)(a^3+b^3+c^3)\le (a^2+b^2+c^2)^3$ For positive real numbers $a,b,c$ prove that $$(a+b+c)(ab+bc+ca)(a^3+b^3+c^3)\le (a^2+b^2+c^2)^3$$
My try :
Rewrite this as $$\frac{a^2+b^2+c^2}{ab+bc+ca}-1\ge \frac{(a+b+c)(a^3+b^3+c^3)}{a^2+b^2+c^2)^2}-1$$
Then I am stumped, I was trying to use SOS but I think an easier proof is possible. My teacher give this to me. He combined to well known inequalities $a^2+b^2+c^2\ge ab+bc+ca$ and $(a+b+c)(a^3+b^3+c^3)\ge (a^2+b^2+c^2)^2$ and changed the sign. This is easy by uvw method
but my teacher wouldn't accept it. Can some one help me? Thanks in advance.
 [/INST] Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$
Hence, we need to prove that $f(w^3)\geq0$, where
$$f(w^3)=(9u^2-6v^2)^3-(27u^2-27uv^2+3w^3)9uv^2$$
and we see that $f$ is a decreasing function, which says
that it's enough to prove our inequality for a maximal value of $w^3$, 
which happens for equality case of two variables.
Since our inequality is homogeneous, we can assume $b=c=1$, which gives
$$(a-1)^2(a^4-2a+4)\geq0,$$
which is obviously true.
Done!
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