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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.

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

1 Answer 1

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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.

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    $\begingroup$ Thanks for the contribution! $\endgroup$
    – user0102
    Aug 6, 2020 at 22:25

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