If $a,b,c$ are non-negative reals such that $a^2+b^2+c^2+abc=4$ , then we have the following nice inequalities:

$\boxed{1} \ \ \ abc \le 1$

Proof: $4=abc+a^2+b^2+c^2 \ge abc+3\sqrt[3]{a^2b^2c^2}$ Let $abc=x^3$ then

$x^3+3x^2 \le 4 \implies (x-1)(x+2)^2 \le 0 \implies x \le 1 \implies abc \le 1$
$\boxed{2} \ \ \ a+b+c \le 3$

Proof(mudok): Two of $(a-1),(b-1),(c-1)$ have the same sign. Assume that $(a-1)(b-1)\ge 0 \ \ \ \iff \ \ \ a+b+c\le 1+c+ab$ . If $c+ab\le 2$ then it is done. If $c+ab>2$ then $4=(a^2+b^2) + (c+ab)c\ge 2ab+(c+ab)c>2ab+2c=2(ab+c)>4$ Contradiction. So $a+b+c\le 3$

$\boxed{3} \ \ \ a+b+c \ge ab+bc+ca$

Proof: $(a+b+c)^2 \ge 3(ab+bc+ca) \ge (a+b+c)(ab+bc+ca) \implies a+b+c \ge ab+bc+ca$

$\boxed{4} \ \ \ a^2+b^2+c^2 \ge 3$

$abc \le 1 \implies 4-(a^2+b^2+c^2) \le 1 \implies a^2+b^2+c^2 \ge 3$

$\boxed{5} \ \ \ a+b+c \ge abc+2$

$\boxed{6}$ (USAMO 2001): $ab+bc+ca \le abc+2$

$\boxed{7}$ (mudok): $2(ab+bc+ca) \le (a+b+c-1)abc+4$

$\boxed{8}$ (own) Refinement of $\boxed3$ : $a+b+c \ge ab+bc+ca+\frac{1}{5}\left[(a-b)^2+(b-c)^2+(c-a)^2\right]$

$\boxed{9}$ (own) Refinement of $\boxed{2+3}$ : $ab+bc+ca+\frac{1}{4}\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \le 3$

$\boxed{10}$ (mudok) $a+b+c \ge ab+bc+ca+\frac{1}{\sqrt{24}}\left[(a-b)^2+(b-c)^2+(c-a)^2\right]$

$\boxed{11}$ $a^2+b^2+c^2+\frac13(ab+bc+ca) \ge 4$

Learning Inequalities