<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 $p=a+b+c, q=ab+bc+ca, r = abc$.  Also note that $p^2\ge 3q$ and $q^2 \ge 3pr$ in general.  WLOG we can set $r = 1$, so $q^2 \ge 3p$ in this case.  The given inequality is then
$$p \cdot q \cdot \left(p(p^3-3q)+3 \right)\le (p^2-2q)^3$$
$$\iff p^6-(7p^4+3p)q +15 p^2 q^2-8 q^3\ge 0$$
$$\iff (p^2-3q)^3+2(p^2-3q)^2q+(q^2-3p)q \ge 0$$
which is obvious using $p^2\ge 3q$ and $q^2 \ge 3p$.
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