From the general form of Adreescu we have: $\frac{a^n}{b+c}+\frac{b^n}{a+c}+\frac{c^n}{a+b}\geq \frac{(a+b+c)^n}{3^{n-2}(2(a+b+c))}=\frac{2^n}{3^{n-2}4S}S^n=(\frac{2}{3})^{n-2}S^{n-1}$ Best Regards, Chris

Here is another solution. $n=1 \: \Longrightarrow$ $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geq \frac{3}{2} \: \Longleftrightarrow $ $(a+b+c)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right) \geq \frac{9}{2}$ which is true by Cauchy-Schwarz inequality. $n \geq 2 \: \Longrightarrow$ By Cauchy-Schwarz inequality again $2(a+b+c)\left(\frac{a^n}{b+c}+\frac{b^n}{c+a}+\frac{c^n}{a+b}\right) \geq \left(a^{\frac{n}{2}}+b^{\frac{n}{2}}+c^{\frac{n}{2}}\right)^2 \geq 3^{2-n}(a+b+c)^n$ $\frac{a^n}{b+c}+\frac{b^n}{c+a}+\frac{c^n}{a+b} \geq \left(\frac{2}{3}\right)^{n-2} S^{n-1}$ Solution is done.

Here is an alternative proof. Note \[\frac{a^{n}}{b+c}+\frac{b^{n}}{c+a}+\frac{c^{n}}{a+b}\geq\left(\frac{2}{3}\right)^{n-2}S^{n-1}\] \[ \iff \frac{a^{n}}{b+c}+\frac{b^{n}}{c+a}+\frac{c^{n}}{a+b}\ge \frac{1}{2\cdot 3^{n-2}}(a+b+c)^{n-1}\] By the power means inequality, $\frac{a^n+b^n+c^n}{3}\ge \left( \frac{a+b+c}{3} \right) ^n$ Assume $a\ge b \ge c$. Then by Chebyshev's inequality $3\left(\frac{a^n}{b+c}+\frac{b^n}{c+a}+\frac{c^n}{a+b}\right) \ge \left(a^n+b^n+c^n\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b} \right)$ Therefore $\frac{a^n}{b+c}+\frac{b^n}{c+a}+\frac{c^n}{a+b}\ge \left(\frac{a^n+b^n+c^n}{3}\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right) $ $\ge \left( \frac{a+b+c}{3} \right) ^n \left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)$ So it suffices to prove $\left( \frac{a+b+c}{3} \right) ^n \left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\ge \frac{1}{2\cdot 3^{n-2}}(a+b+c)^{n-1}$ This rearranges to $(b+c+c+a+a+b)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b} \right)\ge 9$ which is of course the Cauchy-Schwarz Inequality or AM-HM. In fact the conditon $a,b,c$ are sides of a triangle is not necessary, nor is it necessary $n\in \mathbb{N}$, we only need $n\ge 1$ and $n\in \mathbb{R}$.

Letting $a=x+y,b=y+z,c=z+x$ it suffices to prove that: ${\frac{1}{3} *\frac{(x+y)^n}{z}+\frac{1}{3} *\frac{(y+z)^n}{x}+\frac{1}{3} *\frac{(z+x)^n}{y} \geq \left(\frac{2}{3}\right)^{n-2}}(x+y+z)^{n-1}$ Normalizing to $x+y+z=1$ we have from the above inequality ${\frac{1}{3} *\frac{(1-z)^n}{z}+\frac{1}{3} *\frac{(1-x)^n}{x}+\frac{1}{3} *\frac{(1-y)^n}{y} \geq \left(\frac{2}{3}\right)^{n-2}}$ Using Jensen on $f(t)=\frac{(1-t)^n}{t}$ (which is convex for $t$ in the positive reals) the result directly follows. Is this proof legitimate?

We can do it by induction using Rearrangement Inequality as well. (by using the notation from Engel, and extending the scalar product to 3 sequences) The case when $ n = k + 1 $ is similar to the case $n = 1$.

Helder: $(\dfrac{a^{n}}{b+c}+\dfrac{b^{n}}{c+a}+\dfrac{c^{n}}{a+b})((b+c)+(c+a)+(a+b))(1+1+1)(1+1+1)\cdots(1+1+1)\geq (a+b+c)^{n} $ and this is equialent with \[ \frac{a^{n}}{b+c}+\frac{b^{n}}{c+a}+\frac{c^{n}}{a+b}\geq\left(\frac{2}{3}\right)^{n-2}S^{n-1} \]

also $\geq (n-2)(2S-3)$ which is super weak

$a \ge b \ge c \Rightarrow a^{n-1} \ge b^{n-1} \ge c^{n-1}$ and $\frac{a}{b+c} \ge \frac{b}{c+a} \ge \frac{c}{a+b}$ so we have by Tchebycheff: $\sum{\frac{a^n}{b+c}} \ge \frac{(\sum{a^{n-1}})(\sum{\frac{a}{b+c}})}{3}$. ....(1) Now by power-mean inequality: $(\frac{\sum{a^{n-1}}}{3})^{\frac{1}{n-1}} \ge \frac{a+b+c}{3} \Rightarrow \sum{a^{n-1}} \ge \frac{2^{n-1}}{3^{n-2}}S^{n-1}$ Also by the famous Nesbitt inequality: $\sum{\frac{a}{b+c}} \ge \frac{3}{2}$ Combining these inequalities with (1) the result follows.

\begin{align*} \sum_{\text{cyc}}\frac{a^n}{b+c}&\ge\frac{a^n+b^n+c^n}3\sum_{\text{cyc}}\frac1{a+b}\qquad~\text{(Chebyshev)} \\ &\ge\left(\frac{a+b+c}3\right)^n\sum_{\text{cyc}}\frac1{a+b}\qquad\text{(Power Mean)} \\ &=\left(\frac{a+b+c}3\right)^{n-1}\left(\frac{a+b+c}3\sum_{\text{cyc}}\frac1{a+b}\right) \\ &\ge\frac32\left(\frac{a+b+c}3\right)^{n-1}\qquad\qquad~~\text{(Nesbitt)} \\ &=\frac32\cdot s^{n-1}\cdot\left(\frac23\right)^{n-1} \\ &=\left(\frac23\right)^{n-2}s^{n-1}\qquad\square \\ \end{align*}Weird that triangle inequality wasn't used.