How to prove the following inequality
$$(1-x) \ln(1-x) \geq -x + x^2/2\, \textrm{ for } x\in(0,1)?$$
In general how should one go about proving/approximating exponential and logarithms to get the following type of inequalities.
$$ e^{-x} \leq 1 - x + \frac{x^2}2,\; \textrm{ when } x\geq 0 $$
$$ e^{-x} \leq 1 - x/2,\;\; \textrm{ for } x \in [0,1.59] $$
$$ \ln(1+x) \geq x/2,\;\; \textrm{ for } x \in [0,2.5] $$
You could find more of these here.
On
We can rewrite it to $$\cases{f(x) = (1-x)\ln(1-x)+x-\frac{x^2}2\\f(x) \geq 0, x\in[0,1]}$$
Now showing that $f$ is differentiable on the interval is easy so we skip that part.
$$\frac{\partial f}{\partial x}=-\ln(1-x) - \frac{1-x}{1-x} + 1 -x = -x-\ln(1-x)$$ The zeros are where: $$e^x = \frac 1 {1-x}$$ solved by $x=0$ and no other x in the interval $[0,1]$
Now what remains is to calculate $f(0),f(1)$ (or their limits) and draw conclusions.
On
Let $f(x)=\ln(1-x)-\frac{x^2-2x}{2(1-x)}$.
Hence, $f'(x)=\frac{x^2}{2(1-x)^2}>0$, which says $f(x)>f(0)=0$ and we are done!
On
Replacing $x$ by $1-x$ we can see that the inequality is transformed into $$2x\log x\geq x^2-1$$ for $x\in(0,1)$. Consider $$f(x) =2x\log x-x^{2}+1,f(0)=1$$ so that $f$ is continuous on $[0,1]$ and differentiable on $(0,1)$ with derivative $$f'(x) =2+2\log x-2x=2(\log x - x+1)$$ Next we note that $\log x\leq x-1$ so that $f'(x) \leq 0$ and hence $f(x) \geq f(1)=0$ for all $x\in(0,1)$ and we are done.
It is quite trivial that $$ -1-\log(1-t)\geq -1+t \tag{1} $$ holds for any $t\in(0,1)$, hence by considering some $x\in(0,1)$ and applying $\int_{0}^{x}(\ldots)\,dt $ to both sides of $(1)$ the claim follows in a straightforward way.
A much more accurate approximation is $$ \forall x\in(0,1),\qquad (1-x)\log(1-x) \geq \frac{x \left(-6+3 x+2 x^2\right)}{6-x^2}.\tag{2}$$