For $0For $0<t\leq 1$ show
$$ \text{ln}(t)\geq \frac{t-1}{2t+2}\left(1+\sqrt{\frac{2t^2+5t+2}{t}}\right)$$
On a side note, I've been taking this summer class on inequalities. The professor told us that his main source are all sorts of  math olympiads. Given that I'm gonna take Calc 3 next semester and Real Analysis afterwards, are those types of inequalities really as essential as my prof suggests? As you can tell, I'm not really capable of proving most of them by myself yet...

Fact 1: If $x \in (0, 1]$, then
$$\ln x \ge \frac{(x - 1)(1 + x^{1/3})}{x + x^{1/3}}.$$
(See [1], page 272. This inequality is due to Karamata.)
By Fact 1, it suffices to prove that
$$\frac{(t - 1)(1 + t^{1/3})}{t + t^{1/3}}
\ge \frac{t-1}{2t+2}\left(1+\sqrt{\frac{2t^2+5t+2}{t}}\right)$$
or
$$\frac{1 + t^{1/3}}{t + t^{1/3}}
\le \frac{1}{2t+2}\left(1+\sqrt{\frac{2t^2+5t+2}{t}}\right).$$
Letting $t = u^3$, it suffices to prove that, for all $u\in (0, 1]$,
$$\frac{1 + u}{u^3 + u}
\le \frac{1}{2u^3+2}\left(1+\sqrt{\frac{2u^6+5u^3+2}{u^3}}\right)$$
or
$$\frac{2u^4 + u^3 + u + 2}{u^3 + u} \le \sqrt{\frac{2u^6+5u^3+2}{u^3}}$$
or
$$\left(\frac{2u^4 + u^3 + u + 2}{u^3 + u}\right)^2
\le \frac{2u^6+5u^3+2}{u^3}$$
or
$$\frac{2(u^2 + u + 1)(u^2 - u + 1)(1 + u)^2(1 - u)^4}{u^3(u^2 + 1)^2} \ge 0$$
which is true.
We are done.
Reference.
[1] D. S. Mitrinovic, P. M. Vasic, “Analytic Inequalities,” 1970.
