Find infimum of this set

Question

I need to find infimum of this set

$$ \left\{ \frac a {b+c} + \frac b {a+c} + \frac c {a+b} : a,b,c \in \mathbb R^+\right\}$$

Some hint would be helpful

Answer 1

More generally, for all $a,b,c\in\mathbb R^+$, $n\in\mathbb Z^+$:

$$\frac{a}{b+nc}+\frac{b}{c+na}+\frac{c}{a+nb}\ge \frac{3}{n+1}$$

I've posted a proof before in this answer:

First, notice that $a^2+b^2+c^2\ge ab+bc+ca$ for all $a,b,c\in\mathbb R$, because this is equivalent to $\frac{1}{2}\left((a-b)^2+(b-c)^2+(c-a)^2\right)\ge 0$, which is true (or by the Rearrangement Inequality).

Also by the Cauchy-Schwarz Inequality:

$$\sum_{\text{cyc}}\frac{a}{b+nc}\ge \frac{(a+b+c)^2}{\sum_{\text{cyc}}a(b+nc)}=\frac{\sum_{\text{cyc}}a^2+2\sum_{\text{cyc}}ab}{(n+1)\sum_{\text{cyc}}ab}$$

$$\ge \frac{\sum_{\text{cyc}}ab+2\sum_{\text{cyc}}ab}{(n+1)\sum_{\text{cyc}}ab}=\frac{3}{n+1}$$

$n=1$ gives Nesbitt's inequality, which is what you want.

Answer 2

This can be done directly by Nesbitt's Inequality