Because $$\frac{ab+ac+1}{a+2}+\frac{ab+bc+1}{b+2}+\frac{cb+ac+1}{c+2}=3.$$Also, my previous inequality $$\color{blue}{\sqrt{\frac{a+2}{ab+ac+2}}+\sqrt{\frac{b+2}{ab+bc+2}}+\sqrt{\frac{c+2}{cb+ac+2}}\ge \frac{3\sqrt{3}}{2},\ \ \forall a,b,c\ge 0: ab+bc+ca+abc=4.}$$Equality holds iff $a=b=c=1$ or $a=b=\dfrac{4}{3},c=\dfrac{1}{2}$

KhuongTrang wrote: Because $$\frac{ab+ac+1}{a+2}+\frac{ab+bc+1}{b+2}+\frac{cb+ac+1}{c+2}=3.$$ Yes, of course!

arqady wrote: Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc+abc=4.$ . Prove that:

It seems to easy. Is it for junior level?

In fact, it is neither easy,nor solved...

jokehim wrote: It seems to easy. Is it for junior level? I think, it seams easy for you because you saw a solution. During solving this problem you need to prove a new identity and without using computer . It's not so trivial, I think. But you know better and I am wrong, of course.

It is almost obvious when the yield $a+2$ turns. Just remember the identity $$\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=1.$$

jokehim wrote: It is almost obvious when the yield $a+2$ turns. Just remember the identity $$\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=1.$$ I'll say it again. If you know that $\sum_{cyc}\frac{ab+ac}{a+2}=2$ is true, so it's very easy. But this identity is new and for $(2,2,0)$ in the original inequality we obtain something close to equality. All these give a legitimation to try strong methods, which failed or gives something very hard. The competition was in 27.11.2024. I'll say you later what is a number of students(was around 300), which solved this problem. OK?