jred wrote: Let real numbers $a$,$b$,$c$ satisfy $abc=1$. Prove that inequality $\dfrac{a^k}{a+b}+ \dfrac{b^k}{b+c}+\dfrac{c^k}{c+a}\ge \dfrac{3}{2}$ holds for all integer $k$ ($k \ge 2$). By Holder and AM-GM we have: $[(a+b)+(b+c)+(c+a)]3^{k-2}LHS\ge(a+b+c)^k$ $\Longleftrightarrow$ $LHS\ge\frac{(a+b+c)^k}{2(a+b+c)3^{k-2}}\ge\frac{(a+b+c)^{k-1}}{2\cdot3^{k-2}}\ge\frac{(3\sqrt[3]{abc})^{k-1}}{2\cdot3^{k-2}}=\frac{3}{2}$

WLOG we may assume that $a \geq b \geq c$ Then: $\frac{a}{b+c} \geq \frac{b}{a+c} \geq \frac{c}{a+b}$ and $a^{k-1} \geq b^{k-1} \geq c^{k-1}$ By Chebyshev's inequality: $(a^{k-1}+b^{k-1}+c^{k-1})(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}) \leq 3(\frac{a^k}{b+c}+\frac{b^k}{a+c}+\frac{c^k}{a+b})$ By AM-GM: $a^{k-1}+b^{k-1}+c^{k-1} \geq 3\sqrt[3]{(abc)^{k-1}}=3$ By Nesbitt's inequality: $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}$ Using this inequalities we obtain: ${3\frac{a^k}{b+c}+\frac{b^k}{a+c}+\frac{c^k}{a+b}} \geq (a^{k-1}+b^{k-1}+c^{k-1})(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b})\geq3 \cdot \frac{3}{2}$ Therefore: $\frac{a^k}{a+b}+\frac{b^k}{b+c}+\frac{c^k}{c+a}\geq \frac{3}{2} $

97aydos wrote: Using this inequalities we obtain: ${3\frac{a^k}{b+c}+\frac{b^k}{a+c}+\frac{c^k}{a+b}} \geq (a^{k-1}+b^{k-1}+c^{k-1})(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b})=3 \cdot \frac{3}{2}$ Therefore: $\frac{a^k}{a+b}+\frac{b^k}{b+c}+\frac{c^k}{c+a}\geq \frac{3}{2} $ Are you sure