Prove the inequality $\frac{1}{1+a+b}+\frac{1}{1+c+b}+\frac{1}{1+a+c}+a+b+c \le 3 + \frac 13 (ab+bc+ac)$

Question

Let $a,b,c \in [0;1]$. Prove that $$\frac{1}{1+a+b}+\frac{1}{1+c+b}+\frac{1}{1+a+c}+a+b+c \le 3 + \frac 13 (ab+bc+ac).$$

I used Buffalo way. Can anyone offer an easier way

Answer 1

Let $f(a,b,c)=\sum\limits_{cyc}\left(\frac{1}{1+a+b}+a-1-\frac{ab}{3}\right)$.

Hence, $f$ is a convex function of $a$, of $b$ and of $c$.

Thus, $\max\limits_{\{a,b,c\}\subset[0,1]}f=\max\limits_{\{a,b,c\}\subset\{0,1\}}f=f(0,0,0)=0$.

Done!