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...
 A: My try .
here (Which is greater $\frac{13}{32}$ or $\ln \left(\frac{3}{2}\right)$)  Michael Rozenberg show :
let $0<x\leq 1$ :
$$\ln{x}\geq(x-1)\sqrt[3]{\frac{2}{x^2+x}}$$
Now we need to show :
$$(x-1)\sqrt[3]{\frac{2}{x^2+x}}\geq \frac{x-1}{2x+2}\left(1+\sqrt{\frac{2x^{2}+5x+2}{x}}\right)$$
or :
$$\sqrt[3]{\frac{2}{x^2+x}}\geq \frac{1}{2x+2}\left(1+\sqrt{\frac{2x^{2}+5x+2}{x}}\right)$$
I'm stuck here .Hope someone can achieve this .
Edit :
In fact using Wolfram Alpha we can conclude quicly see this logarithmic derivative link
Just some remarks for another proof :
We have the inequality :
$$\sqrt{\frac{\left(2x^{2}+5x+2\right)}{x}}\geq \frac{3}{2}\left(\sqrt{\frac{1}{x^{a}}}+\sqrt{x^{a}}\right)$$
Where :
$$a=\frac{2\sqrt{2}}{3}$$
So we need to show :
$$\left(\frac{3}{2}\left(\sqrt{\frac{1}{x^{a}}}+\sqrt{x^{a}}\right)+1\right)\left(x-1\right)\cdot\frac{1}{\left(2x+2\right)}-\ln\left(x\right)\leq 0$$
Or $x\to x^{\frac{2}{a}}$ :
$$\left(\frac{3}{2}\left(\frac{1}{x}+x\right)+1\right)\left(x^{\frac{2}{a}}-1\right)\cdot\frac{1}{\left(2x^{\frac{2}{a}}+2\right)}-\frac{\ln\left(x\right)2}{a}\leq 0$$
Now we substitute $\ln(x)=y$ and then we can use derivative .
A: Consider only strict inequality by dropping off the equal sign (because equality can be proven easily by putting t = 1).
Let $A$ = ${\frac{(t-1)}{2(t+2)}}$= $(1/2)- 1/(t+1)$
Now, t∈ (0,1) . Hence,
$A = (1/2)- 1/(t+1)∈ (-1/2,0)$
Let $B$  = $\left(1+\sqrt{\frac{2t^2+5t+2}{t}}\right)$
Now, we know that ;
$t + (1/t)  >2$
$2[t + (1/t)] >4$
$2[t + (1/t)] + 5 >9$
$\sqrt{2[t + (1/t)] + 5} >3$
$\sqrt{2[t + (1/t)] + 5} + 1 >4$
Hence,
$B$ $> 4$ because
$\sqrt{2[t + (1/t)] + 5} + 1 = \left(1+\sqrt{\frac{2t^2+5t+2}{t}}\right)$
Hence, your inequality reduces to ;
$ln(t)  > AB$
where $A∈ (-1/2,0)$ and $B ∈ (4,∞)$
Can you proceed further?
A: 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.
