# Prove that $\;0\le-p_1\ln p_1-p_2\ln p_2-\ldots-p_n\ln p_n\le\ln n$

Prove the following property.

If $$\;n\in\mathbb{N}\;,\;p_1,p_2,\ldots,p_n>0\;$$ and $$\;p_1+p_2+\ldots+p_n=1\;,\;$$ then $$0\le-p_1\ln p_1-p_2\ln p_2-\ldots-p_n\ln p_n\le\ln n\;.$$

Could you tell me if my proof is correct?

Is it possible to prove it in a simpler way?

My proof:

Since $$\;f(x)=\ln x:\left]0,+\infty\right[\to\mathbb{R}\;$$ is a concave function, it follows that

$$f(t_1x_1+t_2x_2+\ldots+t_nx_n)\ge t_1f(x_1)+t_2f(x_2)+\ldots+t_nf(x_n)$$

for all $$\;x_1,x_2,\ldots,x_n\in\left]0,+\infty\right[\;$$ and for all $$\;t_1,t_2,\ldots,t_n\ge0\;$$ such that $$\;t_1+t_2+\ldots+t_n=1\;.$$

By letting $$\;x_i=\dfrac{1}{p_i}\in\left]0,+\infty\right[\;,\;t_i=p_i>0\;$$ for any $$\;1\le i\le n\;,\;$$ we get that

$$f\left(p_1\dfrac{1}{p_1}+p_2\dfrac{1}{p_2}+\ldots+p_n\dfrac{1}{p_n}\right)\ge p_1f\left(\dfrac{1}{p_1}\right)+p_2f\left(\dfrac{1}{p_2}\right)+\ldots+p_nf\left(\dfrac{1}{p_n}\right)\;,$$

$$f\left(n\right)\ge p_1f\left(\dfrac{1}{p_1}\right)+p_2f\left(\dfrac{1}{p_2}\right)+\ldots+p_nf\left(\dfrac{1}{p_n}\right)\;.$$

Consequently,

$$\ln n\ge p_1\ln\left(\dfrac{1}{p_1}\right)+p_2\ln\left(\dfrac{1}{p_2}\right)+\ldots+p_n\ln\left(\dfrac{1}{p_n}\right)\;,$$

$$\ln n\ge-p_1\ln p_1-p_2\ln p_2-\ldots-p_n\ln p_n\;,\;$$

$$\color{brown}{-p_1\ln p_1-p_2\ln p_2-\ldots-p_n\ln p_n\le\ln n}\;.$$

Since $$\;p_1,p_2,\ldots,p_n>0\;$$ and $$\;p_1+p_2+\ldots+p_n=1\;,\;$$ it results that $$\;p_1,p_2,\ldots,p_n\in\left]0,1\right]\;,\;$$ hence

$$\color{brown}{-p_1\ln p_1-p_2\ln p_2-\ldots-p_n\ln p_n\ge0}\;.$$

• On taking exp it is just the weighted AM-GM inequality – Albus Dumbledore Nov 28 '20 at 17:09
• Could you write your proof? – Angelo Nov 28 '20 at 17:16
• the best way to prove weighted AM-GM is jensen like you did some other proofs are here: en.wikipedia.org/wiki/…. – Albus Dumbledore Nov 28 '20 at 17:18