<s>[INST] Inequality problem $(a+b+c)^5\geq27(ab+bc+ca)(ab^2+bc^2+ca^2)$ While solving one inequality, I arrived at a much simpler, but still nontrivial inequality
$$(a+b+c)^5\geq27(ab+bc+ca)(ab^2+bc^2+ca^2)$$
where $a,b,c$ are positive real numbers.
It apparently holds, but I can't seem to find a proof.
The problem is it's not symmetric and applying inequalities $(a+b+c)^2\geq3(ab+bc+ca)$ or $a^3+b^3+c^3\geq ab^2+bc^2+ca^2$ won't work.
Any ideas?
 [/INST] Without loss of generality, let $a+b+c=3$. Then we only have to prove
$$(a^2b+b^2c+c^2a)(ab+bc+ac)\le 9$$
Assume $b=\text{mid}(a,b,c)$, then we have
$$c(b-a)(b-c)\le 0\Longleftrightarrow a^2b+b^2c+c^2a\le b(a^2+ac+c^2)$$
So
$$(a^2b+b^2c+c^2a)(ab+bc+ac)\le b(a^2+ac+c^2)(ab+bc+ac)$$
Using AM-GM inequality, we get
$\begin{align*}
b(a^2+ac+b^2)(ab+bc+ca)&\le\,b\cdot\dfrac{(a^2+ac+c^2+ab+bc+ac)^2}{4}=
b\cdot\dfrac{(a+c)^2(a+b+c)^2}{4}\\
&=\dfrac{9}{8}\cdot 2b\cdot(a+c)\cdot(a+c)\\
&\le\dfrac{9}{8}\cdot\dfrac{(2b+a+c+a+c)^3}{27}\\
&=\,9
\end{align*}$
Done!
</s>