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The equation below is true i have proved it using values between 0 and 1 but how can I prove it mathematically? I have already reduced it to its simplest form (according to me) and I am kind of stuck at this point:

$$ H(x)\geq \ln(x)(\ln (1 - x)) \qquad\text{for}\qquad 0\lt x\lt 1$$

where $$ H(x) = -x \ln(x) - (1-x)\ln(1-x)$$ Any help will be quite useful to me. Thanks.

Martin R
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tester124
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3 Answers3

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We have to show that $$ f(x) = x \ln(x) + (1-x) \ln(1-x) + \ln(x) \ln(1-x) \\ = \bigl( \ln(x) + 1 - x \bigr) \bigl( \ln(1-x) + x \bigr) - x(1-x) $$ is $\le 0$ for $0 < x < 1$. Using the “well-known” estimates $$ \frac{x-1}{x} \le \ln(x) \le x-1 \quad (\text{for } x > 0) $$ (see for example How can I prove that $ \frac {x-1}{x}\leq \log x\leq x-1$), the first factor can be estimated as $$ 0 \ge \ln(x) + 1 - x \ge \frac{x-1}{x} + 1-x = -\frac{(1-x)^2}{x} $$ and the second factor as $$ 0 \ge \ln(1-x) + x \ge \frac{-x}{1-x} + x = -\frac{x^2}{1-x} \, . $$ It follows that $$ f(x) \le \frac{(1-x)^2}{x} \cdot \frac{x^2}{1-x} - x(1-x) = 0 \, . $$

Martin R
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Note that for $0\lt x\lt1$, $$ x=\int_0^x1\,\mathrm{d}t\le\overbrace{\int_0^x\frac1{1-t}\,\mathrm{d}t}^{-\log(1-x)}\le\int_0^x\frac1{1-x}\,\mathrm{d}t=\frac{x}{1-x}\tag1 $$ Negating gives, $$ -\frac{x}{1-x}\le\log(1-x)\le-x\tag2 $$ Therefore, $$ -\frac{x^2}{1-x}\le x+\log(1-x)\le0\tag3 $$ and substituting $x\mapsto1-x$, $$ -\frac{(1-x)^2}{x}\le(1-x)+\log(x)\le0\tag4 $$ Multiplying inequalities $(3)$ and $(4)$ gives $$ [x+\log(1-x)][(1-x)+\log(x)]\le x(1-x)\tag5 $$ which upon rearrangement yields $$ \log(1-x)\log(x)\le-x\log(x)-(1-x)\log(x)\tag6 $$ which is the inequality sought.

robjohn
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One approach is to show that the function $$G(x) = \ln(x) \ln(1-x) - H(x)$$ is convex, i.e. $G''(x) \geq 0$ for $0 < x < 1$. Since $G$ is continuous on $[0,1]$ and $G(0) = G(1) = 0$, it follows that $G(x) \leq 0$ on $[0,1]$ yielding the desired inequality.

A direct calculation gives $$ G''(x) = \frac{-\ln(x)}{(1-x)^2} + \frac{-\ln(1-x)}{x^2} - \frac{1}{x(1-x)}. $$

The first two terms have a "good" sign, but the third term has a "bad" sign for $0 < x < 1$. Using the elementary inequalities $-\ln(1-x) \geq x, -\ln(x) \geq 1-x$ for $0 \leq x \leq 1$, we get $ G''(x) \geq 0. $

Shankar Venkataramani
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