Inequality proof $\prod_{cyc} (ab+1)$

Question

Prove that if for positive reals $a,b,c$ with $a^2+b^2+c^2+ab+bc+ca \le 2$ and $a+b+c=1$ then $$(ab+1)(bc+1)(ca+1)\ge ((1-a)(1-b)(1-c))^2.$$

I've tried expanding and i've noticed that $a+b+c=1$ and $a^2+b^2+c^2+ab+bc+ca\le 4$ imply $a^2+b^2+c^2 \le 3$ but i'm not sure how that helps...

Doesn't $a+b+c=1$ and $a,b,c\geq 0$ imply $0\leq a,b,c \leq 1$. Then we have that $(ab+1)(bc+1)(ca+1) \geq 1$ while $0\leq 1-a, 1-b, 1-c \leq 1$ and so the inequality follows naturally? — AspiringMat, Jul 18 '21 at 18:18
If they are all positive real, as per your last edit, then isn't it obvious? All $3$ factors on LHS are $ \gt 1$ and all $3$ factors on RHS are $ \lt 1$ What is left to prove? — Math Lover, Jul 18 '21 at 18:53

score 1 · Answer 1 · answered Jul 18 '21 at 18:38

You don't need the additional constraint $a^2+b^2+c^2+ab+bc+ac\leq 2$.

Expanding on my comment, $a+b+c =1 \land a,b,c\geq 0 \implies 0\leq a,b,c \leq 1$. This also implies that

$$0 \leq 1-a, 1-b, 1-c \leq 1$$

Which further implies that

$$(1-a)(1-b)(1-c) \leq 1$$

Which again implies that

$$((1-a)(1-b)(1-c))^2 \leq 1 \quad (*)$$

However,

$$(1+ab)(1+bc)(1+ac) \geq 1\times 1 \times 1 =1 \quad (**)$$

Combining $*, **$, we get

$$(1+ab)(1+bc)(1+ac) \geq 1 \geq ((1-a)(1-b)(1-c))^2$$

1 Answers1