<s>[INST] Proving the inequality $\frac{9}4\ + \frac{3abc}4\ \ge ab+bc+ca$ If a,b,c are non-negative real numbers for which holds that $a+b+c=3$ then prove the following inequality:  $$\frac{9}4\ + \frac{3abc}4\ \ge ab+bc+ca$$
 [/INST] If we consider the homogeneous form of our inequality:
$$\frac{9}{4}\left(\frac{a+b+c}{3}\right)^3+\frac{3}{4}abc\geq (ab+ac+bc)\left(\frac{a+b+c}{3}\right)$$
we just have to prove that for any triple $(a,b,c)$ of non-negative real numbers
$$ a^3+b^3+c^3+3abc \geq ab(a+b)+bc(b+c)+ac(a+c)\tag{1} $$
holds, but $(1)$ is exactly Schur's inequality.
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