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

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

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


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

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

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

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

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

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

Your proof is correct. I don't see a simpler way than this to prove that.

  • $\begingroup$ Is it possible to prove it by induction or by using derivatives? $\endgroup$ – Angelo Nov 28 '20 at 17:14
  • $\begingroup$ @Angelo It is possible to prove it by induction but it would be the same as proving Jensen's inequality (see here) by induction. And I don't see how derivatives could help here. $\endgroup$ – Michelle Nov 28 '20 at 17:18

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