Example 1.4.1. Let $a, b, c$ be non-negative real numbers. For each real number $k$, find the minimum of the following expression
$$S=\left(\frac{a}{b+c}\right)^{k}+\left(\frac{b}{c+a}\right)^{k}+\left(\frac{c}{a+b}\right)^{k}$$
SOLUTION. Certainly, Nesbitt inequality is a particular case of this inequality for $k=1$. If $k \geq 1$ or $k \leq 0$ then it's easy to deduce that
$$\left(\frac{a}{b+c}\right)^{k}+\left(\frac{b}{c+a}\right)^{k}+\left(\frac{c}{a+b}\right)^{k} \geq \frac{3}{2^{k}}$$

If $k=\frac{1}{2}$, we obtain a familiar result as follows
$$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}} \geq 2$$

The most difficult case is $0 < k < 1$. If $k > \frac{1}{3}$, we have indeed that $h'(u) > 0$. Thus $h(u) \geq h(0) = 0 \Rightarrow g'(u) \geq 0$. Therefore $g$ is a monotonic function if $u \geq 0$ and therefore
$$\sum_{cyc}\left(\frac{a}{b+c}\right)^{k} \geq g(u) \geq g(0) = \frac{2t^k}{(t+c)^k} + \frac{c^k}{(2t)^k} \quad (\star)$$

Because $g(0)$ is homogeneous, we may assume that $c \leq t = 1$. Consider the function
$$p(c) = \frac{2^{k+1}}{(1+c)^k} + c^k$$

Since the derivative
$$p'(c) = \frac{-k \cdot 2^{k+1}}{(1+c)^{k+1}} + k c^{k-1}$$
has the same sign as the function
$$q(c) = (k+1) \ln (c+1) + (k-1) \ln c - (k+1) \ln 2$$
and also because, as it's easy to check
$$q'(c) = \frac{k+1}{c+1} + \frac{k-1}{c} = \frac{(k+1)c + (c+1)(k-1)}{c(c+1)}$$
has no more than one real root, we obtain that $p'(c) = 0$ has no more than one real root in $(0,1)$ (because $p'(1) = 0$). Furthermore, $\lim_{c \to 0} p(c) = +\infty$, so we obtain
$$p(c) \geq \min \left\{p(1); \lim_{c \to 0} p(c)\right\} = \min \left\{3; 2^{k+1}\right\}$$

According to $(\star)$ and $(\star \star)$, we conclude that
$$\left(\frac{a}{b+c}\right)^{k} + \left(\frac{b}{a+c}\right)^{k} + \left(\frac{c}{a+b}\right)^{k} \geq \min \left\{\frac{3}{2^k}; 2\right\}$$

The inequality has been proved in case $k > \frac{1}{3}$, now we will consider the case $k \leq \frac{1}{2}$. Choose three numbers $\alpha, \beta, \gamma$ satisfying that $\sqrt{\alpha} = a^k, \sqrt{\beta} = b^k, \sqrt{\gamma} = c^k$, then
$$(b+c)^{2k} \leq b^{2k} + c^{2k} \Rightarrow \left(\frac{b}{b+c}\right)^{2k} + \left(\frac{c}{c+b}\right)^{2k} \geq 1 \Rightarrow \frac{a^k}{(b+c)^k} \geq \sqrt{\frac{\alpha}{\beta+\gamma}}$$

Constructing similar results and summing up, we get
$$\frac{a^k}{(b+c)^k} + \frac{b^k}{(c+a)^k} + \frac{c^k}{(a+b)^k} \geq \sqrt{\frac{\alpha}{\beta+\gamma}} + \sqrt{\frac{\beta}{\alpha+\gamma}} + \sqrt{\frac{\gamma}{\alpha+\beta}} \geq 2.$$

Therefore the problem has been completely solved, with the conclusion that
$$\left(\frac{a}{b+c}\right)^{k} + \left(\frac{b}{a+c}\right)^{k} + \left(\frac{c}{a+b}\right)^{k} \geq \min \left\{\frac{3}{2^k}; 2\right\}$$

If $k = \frac{\ln 3}{\ln 2} - 1$, the equality holds for $a = b = c$ and $a = b, c = 0$ up to permutation. Otherwise, the equality only holds in the case $a = b = c$.

In volume I we have a problem from the Vietnam TST 2006, where we have already proved that if $a, b, c$ are the side-lengths of a triangle then
$$(a+b+c)\left(\frac{1}{a} + \frac{a}{b} + \frac{1}{c}\right) \geq 6\left(\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}\right)$$
or, in other words, $N(a, b, c) \geq 3 N(a+b, b+c, c+a)$. Moreover, we also have some other nice results related to the expression $N$ as follows