# Question about convexity: how do we prove that $\displaystyle \sum_{i=1}^{k}p_{i}b_{i}\geq\prod_{i=1}^{k}b^{p_{i}}_{i}$?

Let $$b_{1},b_{2},\ldots,b_{k}$$ be nonnegative numbers and $$p_{1} + p_{2} + \ldots + p_{k} = 1$$ where each $$p_{i}$$ is positive. Then

\begin{align*} \sum_{i=1}^{k}p_{i}b_{i}\geq\prod_{i=1}^{k}b^{p_{i}}_{i} \end{align*}

MY ATTEMPT

Since the logarithm function is strictly increasing, the proposed inequality is equivalent to \begin{align*} \ln\left(p_{1}b_{1} + p_{2}b_{2} + \ldots + p_{k}b_{k}\right) \geq p_{1}\ln(b_{1}) + p_{2}\ln(b_{2}) + \ldots + p_{k}\ln(b_{k}) \end{align*}

Once $$f''(x) < 0$$, where $$f(x) = \ln(x)$$, we conclude that $$f$$ is concave and the proposed inequality holds.

My question is: am I proving this result correctly? If this is the case, is there another way to prove it?

Any contribution is appreciated.

• You have done it absolutely correctly. Another method is by using weighted AM-GM as shown in my answer that I'm typing. Aug 6, 2020 at 22:17

Using $$b_1,b_2,\cdots,b_k\ge 0$$ with $$p_1,p_2,\cdots,p_k$$ as the respective weights, weighted AM-GM inequality gives $$\frac{\sum_{i=1}^k p_ib_i}{\sum_{i=1}^k p_i} \ge \left(\prod_{i=1}^k b_i^{p_i}\right)^{\dfrac{1}{\sum_{i=1}^k p_i}}$$ which gives the required inequality.
Note: This is not much different from your approach because the special case of Jensen's inequality for the function $$f(x)=\ln(x)$$ can be proved by weighted AM-GM or vice-versa, the weighted AM-GM inequality can be proved by the Jensen's inequality for $$f(x)=\ln(x)$$ as you have done, since the inequality you want to prove is directly the weighted AM-GM inequality in disguise. Also, these two can be proven independently without using each other.