TITLE: Proving $x \ln^2 x - (x-1)^2<0$ for all $x\in(0,1)$
QUESTION [5 upvotes]: I want to prove that for all $x\in(0,1)$,$$f(x):=x \ln^2 x - (x-1)^2<0$$
Using the derivative ($f'(x)=-2x+\ln^2 x+2\ln x +2$), I tried to prove that $f$ is monotonically increasing in $(0,1)$, and then use the fact that $f(1)=0$, but I find the derivative itself complicated. Any suggestions?

REPLY [3 votes]: With the change of variable $x=e^{-t}$ the inequality turns out to be equivalent to:
$$ \forall t\in\mathbb{R}^+,\qquad t^2 e^{-t} \leq (1-e^{-t})^2 $$
or to:
$$ \forall t\in\mathbb{R}^+,\qquad t^2 \leq (2\sinh(t/2))^2 $$
but:
$$ \forall t\in\mathbb{R}^+, \qquad 2\sinh\frac{t}{2}\geq t $$
is trivial since $\sinh(y)$ is convex over $y>0$.

REPLY [2 votes]: $f(x)=x\ln^2x-(x-1)^2$. Then, we have
$$f'(x)=\ln^2x+2\ln x-2(x-1)$$
$$f''(x)=\frac{2\ln x}{x}+\frac 2x-2=\frac{2(\ln x+1-x)}{x}$$
Now let $g(x)=\ln x+1-x$. Then, we have
$$g'(x)=\frac 1x-1=\frac{1-x}{x}\gt 0.$$
Hence, $g(x)$ is strictly increasing for $0\lt x \lt 1$, and with $g(1)=0$, we know that $g(x)\lt 0$ for $0\lt x\lt 1$. Hence, we know that $f''(x)\lt 0$. It follows that $f'(x)$ is strictly decreasing, and with $f'(1)=0$, we know that $f'(x)\gt 0$ for $0\lt x\lt 1$. 
Hence, we know that $f(x)$ is strictly increasing for $0\lt x\lt 1$, and with $f(1)=0$, we know that $f(x)\lt 0$ for $0\lt x\lt 1$.