How to prove $x\ln^2(x)\leq(x-1)^2$ for every real number $x$ I really need help with this proof for my analysis class. I have to prove the following inequality : 
$$(\forall x \in \mathbb R^*_+) \qquad \quad x\cdot \ln^2(x)\leq(x-1)^2$$
I tried, with a simple calculus approach, to compute $f'(x)=[(x-1)^2-x \cdot \ln^2(x)]'$, yet it is too long and I guess it is not the best way to prove it. ( I have to compute $f''(x)$ and $f'''(x)$). Any help ?
 A: Since our inequality does not depend on replacing $x$ on $\frac{1}{x},$ 
it's enough to prove that $f(x)\geq0,$ for $x\geq1$, where
$$f(x)=\sqrt{x}-\frac{1}{\sqrt{x}}-\ln{x}.$$
Indeed, $$f'(x)=\frac{(\sqrt{x}-1)^2}{2\sqrt{x^3}}\geq0,$$ which says $$f(x)\geq f(1)=0$$ and we are done.
A: Let $f(x) = x (\ln x)^2 - (x-1)^2$.
By inspection, note that $f'(x)$ is positive for $0 < x <1$, zero at $x=1$, and negative on $x > 1$.
This tells us that $f(x) \leq f(1)$.
Compute $f(1)$ via L'hopitals rule to find $f(1)=1$. Thus we have our bound.
A: Note, for any $x>0$,
$$(x-1)^2-x\ln^2 x$$
$$=x(\sqrt x-\frac1{\sqrt x} -\ln x)(\sqrt x-\frac1{\sqrt x} +\ln x)$$
$$=x\int_1^x d\left(\sqrt t-\frac1{\sqrt t} -\ln t\right)
\cdot\int_1^x d\left(\sqrt t-\frac1{\sqrt t}+\ln t\right)$$
$$=\frac14 x\int_1^x \frac{(1-t^{1/2})^2}{t^{3/2}}dt
\cdot\int_1^x \frac{(1+t^{1/2})^2}{t^{3/2}}dt\ge0$$
The inequality  holds because the two integrals have the same sign.
A: If
$$
g(x) = \frac{(x - 1)^2}{x} \quad (x > 0),
$$
then
$$
g\left(\frac1x\right) = g(x) \quad (x > 0).
$$
But also
$$
\ln\left(\frac1x\right)^2 = \ln(x)^2 \quad (x > 0),
$$
therefore it is enough to prove
$$
\ln(x)^2 < \frac{(x - 1)^2}{x} \quad (x > 1).
$$
Equivalently:

$$
\ln(x) < \frac{x - 1}{\sqrt{x}} \quad (x > 1).
$$

This follows from the convexity of the logarithm function.
Proof. Write $x = t^2,$ where $t > 1.$ Then, because the graph of a convex function lies below any of its chords,
$$
\ln(x) = 2\ln(t) = 2\int_1^t\frac{du}u < 2(t - 1) \cdot \frac12\left(1 + \frac1t\right)
= \frac{t^2 - 1}t = \frac{x - 1}{\sqrt{x}}.
$$
