cjquines0 wrote: Let $k$ be a positive real number such that $$\frac{1}{k+a} + \frac{1}{k+b} + \frac{1}{k+c} \leq 1$$for any positive positive real numbers $a$, $b$ and $c$ with $abc = 1$. Find the minimum value of $k$. $k_{min}=2.$

nice inequality with abc=1: Let $ a,b,c> 0 $ and $ abc=1 $. Prove that\[ \frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}\leq \frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}\le 1. \]here Let $ a,b,c> 0 $ and $ abc=1 $. Prove that\[ \frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}\le 1. \]Proof: $\sum_{cyc} \frac{a}{2+a}=\sum_{cyc} \frac{a^2}{2a+a^2}\geq\frac{(a+b+c)^2}{2(a+b+c)+a^2+b^2+c^2}\geq 1$, $\sum_{cyc} \frac{1}{2+a}=\frac{3}{2}-\frac{1}{2}\sum_{cyc} \frac{a}{2+a}\leq 1.$

sqing wrote: $\sum_{cyc} \frac{a^2}{2a+a^2}\geq\frac{(a+b+c)^2}{2(a+b+c)+a^2+b^2+c^2}\geq 1$ Could you explain this in detail?