<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] We need to show that for $t> 0$,
$$4\ln t \ln (t + 2) + 4 \ln (t + 1) \ln (t + 3) > \ln t \ln ( t + 1) + 3\ln t \ln (t + 3) + 3\ln (t + 1) \ln (t + 2) $$
Case $t> 1$ 
Note that $\ln t, \ln (t+1)$ and $\ln (t+2), \ln(t+3)$ are similarly ordered, so by Rearrangement 
$$\ln t \ln(t+2)+\ln (t+1) \ln(t+3) > \ln t \ln(t+3) + \ln (t+1) \ln (t+2)$$
So it is enough to show that
$$\ln t \ln (t + 2) +  \ln (t + 1) \ln (t + 3) > \ln t \ln ( t + 1) $$
$$\iff \ln t(\ln (t + 2)-\ln(t+1)) + \ln(t+1)\ln (t + 3) > 0$$
which is obvious as $\ln$ is increasing.
Case $0< t \le 1$
We can write the inequality as
$$\ln (t + 1) \ln \frac{(t+3)^4}{(t+2)^3} > \ln t \ln \frac{(t+1)(t+3)^3}{(t+2)^4}$$
As the fractions in the arguments are $> 1$, the LHS is clearly positive while the RHS is negative from the $\ln t$ term.
</s>