If $a, b, c > 0$ and $n \in \mathbb{N}^*$, such that $a^n+b^n+c^n=3$ it seems true that $$ \frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} \geq 3. $$
Have a counterexample if it is false for some $n$.
Let $$a=1,b=\sqrt[8]{\frac{1}{2}}, c=\sqrt[8]{2-\frac{1}{2}}$$
Than $a^8+b^8+c^8=3$ and $$ \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\approx 2.9965288195820197 $$ As shown here the inequality $$ \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}=2^{\frac{1}{x}}+6^{\frac{1}{x}}+\left(\frac{3}{2}\right)^{\frac{2}{x}}<3 $$ is always true for $x>7$. So $$ a=1, b=2^{-\frac{1}{n}}, c=\left(\frac{3}{2}\right)^{\frac{1}{n}} $$ are a counterexample for any $n \ge 8$
For $n=7$ we can choose $$ a=1, b=\sqrt[7]{\frac{3}{4}}, c=\sqrt[7]{\frac{5}{4}} $$ and we have $$ \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\approx 2.9999784302204775 $$ (similarly as above this leads to a counterexample for $n \ge 7$, see here )
