Prove that for all $x>0$, $1+2\ln x\leq x^2$ Prove that for all $x>0$, 
$$1+2\ln x\leq x^2$$
How can one prove that? 
 A: First, note the problem cannot be for all $\;x\in\Bbb R\;$ but, at most , for$\;x>0\;$ (why?). Now, a simple calculus approach: define
$$f(x):=1+2\log x-x^2\implies f'(x)=\frac2x-2x=0\iff x=1$$
and since $\;f''(1)=-2<0\;$ , we get a maximum point at $\;(1,0)\;$ , meaning that for all $\;x<0\;$ we have
$$f(x)\le 0$$
A: Notice by the famous inequality $e^x \geq 1 + x $. We can rescale this and obtain
$$ e^{x-1} \geq x \implies x-1 \geq \ln x\implies2x -2 \geq 2\ln x$$
$$ \implies 2x - 1 \geq 2 \ln x + 1 $$
Now, by using $\frac{a+b}{2} \geq \sqrt{ab}$ with $a = x^2 $ and $b = 1$, we obtain
$$ \frac{x^2 + 1}{2} \geq x \implies x^2 \geq 2x - 1$$
Now, by transitivity, $$ x^2 \geq 2 \ln x + 1 $$
QED
Actually a shorter solution:
Since $e^x \geq 1 + x $, then $e^{x^2} \geq 1 + x^2$. Translate this to obtain
$$ e^{x^2-1} \geq x^2 \implies x^2 -1 \geq \ln x^2 \implies x^2 \geq 1 + 2 \ln x$$
A: $\textbf{Hint:}$ Find the minimum value of $f(x) = x^2 - 1 - 2 \ln x$. If it is non-negative, then $x^2 \geq 1 + 2 \ln x$ for all $x$ in the domain.
A: For all $x > 0$, $\ln(x) < x - 1$.
So $1 + 2\ln(x) = 1 + \ln(x^2) \leq 1 + x^2 - 1 = x^2$.
A: I hope you are aware of the standard inequality $e^{x} > 1 + x$ for all $x \neq 0$. This translates to $\log x < x - 1$ for all $x > 0$ and $ x \neq 1$. Now it is easy to see that $1 + 2\log x < 1 + 2(x - 1) = 2x - 1$ and this will be less than equal to $x^{2}$ if $0 \leq x^{2} - 2x + 1 = (x - 1)^{2}$ which is true. So we have $1 + 2 \log x \leq x^{2}$ for all $x > 0$. Equality occurs when $x = 1$.
A: Hint: $ O(c + \log n) \subset O(n^2) $.
All logarithms grow asymptotically slower than all polynomials. You can use this theorem to prove that the inequality holds for the constant $ k = 1 $ and holds for arbitrarily small $x$.
It's a kind of induction.
PS. If you are unfamiliar with  Big-O notaiton, the definition is:
$$ O(g) = \{ f \mid \forall k > 0\; \exists n > 0\; \forall x > n : |f(x)| \le |k\,g(x)| \} $$
Meaning, $f$ is in $O(g)$ if an only if $f$ is after some point always smaller than $g$ times some constant.
