Cyclic inequality with a+b+c=3 Let $a;b;c$ be non-negative real numbers such that $a+b+c=3$
Prove:
$(a+b)(b+c)(c+a) \geq (a+bc)(b+ca)(c+ab)$
Can someone prove that problem without using $pqr$ or $uvw$ method? Thanks so much.

Using the AM-GM inequality, we have
$$(a + bc)(b + ca) \leqslant \left[\frac{(a + bc)+(b + ca)}{2}\right]^2 = \frac{(a + b)^2(c + 1)^2}{4} .$$
Therefore
$$(a+bc)(b+ca)(c+ab) \leqslant \frac{\displaystyle \prod (a+b) \prod (a+1)}{8}.$$
It remains to prove that
$$(a+1)(b+1)(c+1) \leqslant 8.$$
Which is true because
$$(a+1)(b+1)(c+1) \leqslant \left(\frac{a+1+b+1+c+1}{3}\right)^3=8.$$
Done.
