how to solve the following inequality

I am stuck on the following problem that says:

If $a,b,c$ are three positive numbers in harmonic progression, prove that $$a^n+c^n>2b^n, n>1$$, $n$ is a natural number.

My try: Since $a,b,c$ are three positive numbers in harmonic progression, $$\frac1a,\frac1b,\frac1c$$ are in arithmetic progression that gives $$\frac1a+\frac1c=\frac2b \implies b(a+c)=2ac.$$

Also, we know that if $a_1,a_2,a_3,...........,a_n$ are $n$ positive real numbers , not all equal , and $m$ be a rational number,then $$\frac{a_1^m+a_2^m+a_3^m+...........+a_n^m}{n}> \text{or}< \left(\frac{a_1+a_2+a_3+...........+a_n}{n}\right)^m$$ according as $m$ does not or does lie between $0$ and $1$

And so, $$\frac{a^n+c^n}{2}>\left(\frac{a+c}{2}\right)^n=\left(\frac{2ac}{b}\right)^n$$..

It's $$a^n+c^n\geq2\left(\frac{2}{\frac{1}{a}+\frac{1}{c}}\right)^n$$ or $$(a^n+c^n)\left(\frac{a+c}{2}\right)^n\geq2a^nc^n,$$ which is true by AM-GM: $$a+c\geq2\sqrt{ac}$$ and $$a^n+c^n\geq2\sqrt{a^nc^n}.$$