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?
Let $a,b,c,x,y,z&gt;0$ and $x+ y+ z= 1$. Show that $$ax+ by+ cz\geq  a^{x}.b^{y}.c^{z}$$ I have found the generalization of this inequality: For 2 sets of numbers $\left \{ a_{i} \right \}, \left \{ b_{i} \right \}&gt; 0$ and $\sum_{1}^{n}b_{i}= 1$, we have: $$\sum_{1}^{n}\left ( a_{i}b_{i} \right )\geq  \prod_{1}^{n}\left ( a_{i}^{b_{i}} \right )$$ First solution Simply compare the logs: as the natural logarithm is a concave function, one has $$\ln\bigl(a^x b^y c^z)=x\ln a+y\ln b+z\ln c\le \ln(ax+by+cz)$$ for all non-negative $x,y,z$ such that $x+y+z=1$. How about another solution? I really hope to see that. Thanks!
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So far in the book, Rudin has not discussed the functions $\cos$ and $\sin$. So can we come up with an example that uses only what Rudin has discussed? Central limit theorem and the law of large numbers I have a very beginner's question regarding the Central Limit Theorem (CLT):  I am aware that the CLT states that a mean of i.i.d. random variables is approximately normal distributed (for $n \to \infty$, where $n$ is the index of the summands) or the standardized random variable would have a standard normal distribution.  Now the Law of Large Number states roughly speaking that the mean of i.i.d random variables converges (in probability or almost surely) to their expected value.  What I don't understand is: If, as the CLT states, the mean is approximately normally distributed, how then can it also converge to the expected value at the same time?   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