# Prove the following inequality??

Prove that $$(1+k)^{p}<\frac{(1-k^{p+1})}{(1-k)},$$ Here k,p $\in$ (0,1). I have proved this ineqality for p=1/2. The proof is as follows. \begin{eqnarray} (1+k)^{0.5}<\frac{(1-k^{1.5})}{(1-k)} \\ (1+k)^{0.5}<\frac{(1-\sqrt{k})(k+1+\sqrt{k})}{(1-\sqrt{k})(1+\sqrt{k})}\\ (1+k)^{0.5}(1+\sqrt{k})<(k+1+\sqrt{k})\\ \text{Geometric Mean}<\text{Arithmatic Mean} \end{eqnarray}, However I am not able to show this inequality for p $\in$ (0,1) with k lying between 0 to 1. I have checked the inequality to hold numerically. Please have a look at it.

• have you heard about Bernoulli's inequality? – Vasya Apr 12 '18 at 17:32
• Thanks for your comment. I saw the inequality right now, don't know if it could be used here somehow. Do you think it would be helpful – Rahul Sharma Apr 12 '18 at 17:48

## 1 Answer

You can indeed show it using Bernouilli's inequality. It states that for $x > -1$ and $0 < r < 1$, $(1 + x)^r < 1 + rx$. The given inequality can be rewritten as $$(1 - k)(1 + k)^p < 1 - k^{p + 1}.$$ Applying Bernouilli's inequality to the LHS, we get $$(1 - k)(1 + k)^p < (1 - k)(1 + pk) = 1 - k - pk^2 + pk.$$ If we can show that this is strictly less than $1 - k^{p + 1}$ or, equivalently, that $$k^p < 1 + pk - p,$$ we are done. This follows however from $k^p = (1 + (k - 1))^p < 1 + p(k - 1)$ by Bernouilli's inequality.