How to prove $x\ln^2(x)\leq(x-1)^2$ for every real number $x$

Question

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 ?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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, $$ \ln(x) = 2\ln(t) = 2\int_1^t\frac{du}u < 2(t - 1) \cdot \frac12\left(1 + \frac1t\right) = \frac{t^2 - 1}t = \frac{x - 1}{\sqrt{x}}. $$