Question: Let $a_1,a_2,\cdots,a_n$ be positive real numbers and $\lambda_1,\lambda_2,\cdots,\lambda_n$ be non-negative real numbers such that $$\sum_{k=1}^n\lambda_k=1.$$ Prove that $$\prod_{i=1}^n a_i^{\lambda_i}\le \sum_{i=1}^n\lambda_i a_i.$$
Solution: Observe that for each $a_i$, $1\le i\le n$, we have an unique $y_i\in\mathbb{R}$, such that $a_i=e^{y_i}$. Therefore, we have $$a_1=e^{y_1},a_2=e^{y_2},\cdots, a_n=e^{y_n}.$$
Discrete Jensen Inequality: Let $f$ be a convex function defined on an interval $I\in\mathbb{R}$. Assume that $x_1,x_2,\cdots,x_n\in I$ and let $\lambda_1,\lambda_2,\cdots, \lambda_n$ be non-negative real numbers such that $\sum_{k=1}^n\lambda_k=1$. Then, $$f\left(\sum_{i=1}^n \lambda_i x_i\right)\le \sum_{i=1}^n\lambda_if(x_i).$$
Now $f(x):=e^x, \forall x\in\mathbb{R}$ is a convex function. Also we have $a_i=e^{y_i}$, $1\le i\le n$, where each $y_i\in\mathbb{R}$ and $\lambda_1,\lambda_2,\cdots,\lambda_n$ are given non-negative real numbers such that $\sum_{k=1}^n\lambda_k=1$. Thus, applying Discrete Jensen Inequality we have $$e^{\sum_{i=1}^n\lambda_i y_i}\le \sum_{i=1}^n\lambda_i e^{y_i}\\\implies \prod_{i=1}^n (e^{y_i})^{\lambda_i}\le \sum_{i=1}^n\lambda_i e^{y_i}\\\implies \prod_{i=1}^n a_i^{\lambda_i}\le \sum_{i=1}^n\lambda_i a_i.$$ Hence, we are done.
Is this solution correct and rigorous enough? And, is there any other way to solve the same?