Let define $$H(x)=x\log x+(1-x)\log(1-x)$$ and $$H(y) = y\log(y) + (1-y)\log(1-y)$$ and $$H(xy) = xy\log xy + (1-xy)\log (1-xy)$$ Here $0\le x,y \le 1$. How can I prove that $H(x)+H(y) \le H(xy)$. Thanks.
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1Take a look here:http://math.stackexchange.com/questions/1897698/proving-entropy-inequalities – Robert Z Sep 14 '16 at 10:02
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1Thanks. But I think that is not a proof. – Parveen Kumar Sep 14 '16 at 13:35
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Fix $y \in (0,1)$ and define
$$ f(x) = H(xy) - H(x) - H(y) $$ With some algebra you can verify that $$ f'(x) = \log \left (\frac{1-x}{x}\right) - y \log \left( \frac{1-xy}{xy} \right) $$
$$ f''(x) = \frac{-1}{x(1-x)} + \frac{y^2}{(1-xy)(xy)} < 0 $$ This means $f'$ is a decreasing function. When $x \rightarrow 0^+$ and $x \rightarrow 1^-$ we have $\ f' \rightarrow \infty$ and $f' \rightarrow -\infty$. So, by IVT we know that there exits $x_0 \in (0,1)$ such that $$ \log \left (\frac{1-x_0}{x_0}\right) - y \log \left( \frac{1-x_0y}{x_0y} \right)=0 $$ So $f(x)$ is increasing on $(0,x_0)$, decreasing on $(x_0,1)$ and achieves its maximum at $x_0$.It is not hard to see that $$ f(0^+) = - H(y) > 0 $$ $$ f(1^-) = - H(1^-) = 0 $$ Hence $f$ should be nonnegative on $(0,1)$.
iamvegan
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