0
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.