<s>[INST] A unusual inequality about function $\ln$ These day,I met a unusual inequality when I solve a difficult problem, and proving the inequality means I have done the work!
Could you show me how to prove it or deny it? By the way, I believe that it's true!

Prove that, for all $t > 0$,
\begin{align*}
&4\ln t\ln (t + 2) - \ln t\ln (t + 1) - 3\ln t\ln (t + 3)\\
 + &4\ln (t + 1)\ln (t + 3) - 3\ln (t + 1)\ln (t + 2) - \ln (t + 2)\ln \left( {t + 3} \right)>0.
\end{align*}

Let
$$f\left( t \right) = 4\ln t\ln \left( {t + 2} \right) - \ln t\ln \left( {t + 1} \right) - 3\ln t\ln \left( {t + 3} \right) + 4\ln \left( {t + 1} \right)\ln \left( {t + 3} \right) - 3\ln \left( {t + 1} \right)\ln \left( {t + 2} \right) - \ln \left( {t + 2} \right)\ln \left( {t + 3} \right),$$
We have
$$f'\left( t \right) = \frac{{2\left[ {{t^2}\ln t - 3{{\left( {t + 1} \right)}^2}\ln \left( {t + 1} \right) + 3{{\left( {t + 2} \right)}^2}\ln \left( {t + 2} \right) - {{\left( {t + 3} \right)}^2}\ln \left( {t + 3} \right)} \right]}}{{t\left( {t + 1} \right)\left( {t + 2} \right)\left( {t + 3} \right)}}.$$
Let
$$g\left( t \right) = {t^2}\ln t - 3{\left( {t + 1} \right)^2}\ln \left( {t + 1} \right) + 3{\left( {t + 2} \right)^2}\ln \left( {t + 2} \right) - {\left( {t + 3} \right)^2}\ln \left( {t + 3} \right),$$
we got
$$g'\left( t \right) = 2\left[ {t\ln t - 3\left( {t + 1} \right)\ln \left( {t + 1} \right) + 3\left( {t + 2} \right)\ln \left( {t + 2} \right) - \left( {t + 3} \right)\ln \left( {t + 3} \right)} \right].$$
And let
$$h\left( x \right) = t\ln t - 3\left( {t + 1} \right)\ln \left( {t + 1} \right) + 3\left( {t + 2} \right)\ln \left( {t + 2} \right) - \left( {t + 3} \right)\ln \left( {t + 3} \right),$$
we have
\begin{align*}
h'\left( x \right) &= \ln t - 3\ln \left( {t + 1} \right) + 3\ln \left( {t + 2} \right) - \ln \left( {t + 3} \right)\\
&= \ln \frac{{t{{\left( {t + 2} \right)}^3}}}{{{{\left( {t + 1} \right)}^3}\left( {t + 3} \right)}} = \ln \left[ {1 - \frac{{2t + 3}}{{{{\left( {t + 1} \right)}^3}\left( {t + 3} \right)}}} \right] < 0.
\end{align*}
However, it seems that there are no use!
 [/INST] The desired inequality is written as
$$3ac - ab - b^2 - bc \ge 0$$
where
$a = \ln (t + 1) - \ln t, ~
b = \ln(t + 2) - \ln(t + 1)$ and
$c = \ln(t + 3) - \ln(t + 2)$.
Clearly, $a > b > c > 0$.
Fact 1: It holds that, for all $u \ge 0$,
$$\frac{2u}{2 + u} \le \ln(1 + u) \le \frac{u}{2}\cdot \frac{2 + u}{1 + u}.$$
(Note: See bounds for logarithm. It is also easily proved by taking derivative.)
Using Fact 1, we have
$$c \ge \frac{2}{2t + 5}, \quad a \ge \frac{2}{2t + 1}, \quad b \le \frac{2t + 3}{2(t + 1)(t + 2)}. \tag{1}$$
(Note: For example, we have $a = \ln(1 + \frac{1}{t})$. Let $u = \frac{1}{t}$.)
Using (1) and $a > b > 0$, we have
\begin{align*}
 & 3ac - ab - b^2 - bc\\
 =\,& (3a - b) c - ab - b^2\\
 \ge\,& (3a - b)\cdot \frac{2}{2t + 5} - ab - b^2 \\
 =\,& \left(\frac{6}{2t + 5} - b\right)a - \frac{2b}{2t+5} - b^2 \\
 \ge\,& \left(\frac{6}{2t + 5} - \frac{2t + 3}{2(t + 1)(t + 2)}\right)a - \frac{2b}{2t+5} - b^2\\
 =\,& \frac{8t^2 + 20t + 9}{2(2t+5)(t+1)(t+2)} a - \frac{2b}{2t+5} - b^2\\
 \ge\,& \frac{8t^2 + 20t + 9}{2(2t+5)(t+1)(t+2)} \cdot \frac{2}{2t+1} - \frac{2}{2t+5}\cdot \frac{2t+3}{2(t+1)(t+2)} - \left(\frac{2t+3}{2(t+1)(t+2)}\right)^2\\
 =\,& \frac{3}{4(2t+5)(t+1)^2(t+2)^2(2t+1)}\\
 >\,& 0.
\end{align*}
We are done.
</s>