If $a, b, c$ are in H.P and $n > 1$ , show that $a^{n} + c^{n} > 2 b^{n}$ .

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Solution

Let $a^{n} a n d c^{n}$ be two numbers

$A M > G M$

then $\frac{a^{n} + c^{n}}{2} > {(a^{n} . c^{n})}^{\frac{1}{2}}$

$a^{n} + c^{n} > 2 {(a . c)}^{\frac{n}{2}}$ ---- (1)

Also $G . M . > H . M .$

i.e. $\sqrt{a c} > b {a c}^{\frac{n}{2}} > b^{n}$ ---- (2)

Hence from $(1)$ and $(2)$ $a^{n} + c^{n} > 2 b^{n}$

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