<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] It's not as hard as it seems (if we use Maple, of course :).
Let's rewrite our inequality:
$$4\ln(t)\ln(t+2)+4\ln(t+1)\ln(t+3) > \\\ln(t)\ln(t+1)+3\ln(t)\ln(t+3)+3\ln(t+1)\ln(t+2)+\ln(t+2)\ln(t+3)$$
Now we use $\ln(t+1)=\ln t+\ln(1+\frac1t)$ and denote $x=1/t$. LHS is
$$4\ln(t)\cdot\big(\ln(t)+\ln(1+2x)\big)+4\big(\ln(t)+\ln(1+x)\big)\cdot\big(\ln(t)+\ln(1+3x)\big),$$
or
$$8\ln^2(t)+4\ln(t)\ln(1+2x)+4\ln(t)\ln(1+3x)+4\ln(1+x)\ln(t)+4\ln(1+x)\ln(1+3x).$$
RHS is
$$\ln(t)\big(\ln(t)+\ln(1+x)\big)+3\ln(t)\big(\ln(t)+\ln(1+3x)\big)+\\+3\big(\ln(t)+\ln(1+x)\big)\big(\ln(t)+\ln(1+2x)\big)+\big(\ln(t)+\ln(1+2x)\big)\big(\ln(t)+\ln(1+3x)\big),$$
or
$$8\ln^2(t)+4\ln(1+x)\ln(t)+4\ln(t)\ln(1+3x)+4\ln(t)\ln(1+2x)+\\+3\ln(1+x)\ln(1+2x)+\ln(1+2x)\ln(1+3x).$$
We see thar $\mathrm{LHS} - \mathrm{RHS}>0$ is just
$$h(x) = 4\ln(1+x)\ln(1+3x)-3\ln(1+x)\ln(1+2x)-\ln(1+2x)\ln(1+3x)>0.$$
Now we will use that: if $f(0)=0$, $f'(x)>0$ for $x>0$, then $f(x) > 0$ for $x>0$. Our goal is to show that $h(x) > 0$ for $x>0$ ($x=1/t>0$). Let's go.
$h(0)=0$, and
$$h'(x) = \frac{1}{(1+x)(1+2x)(1+3x)}\times\\\times \left( \left ( 6\,{x}^{2}+6+12\,x \right) \ln  \left( 1+x \right) + \left( -
24\,x-24\,{x}^{2}-6 \right) \ln  \left( 1+2\,x \right) + \left( 2+12\,
x+18\,{x}^{2} \right) \ln  \left( 1+3\,x \right) \right)$$
or
$$h'(x) = \frac{2}{(1+x)(1+2x)(1+3x)}\times\\\times \left(6\, \left( 1+x \right) ^{2}\ln  \left( 1+x \right) -6\, \left( 1+2\,x
 \right) ^{2}\ln  \left( 1+2\,x \right) +2\, \left( 1+3\,x \right) ^{2
}\ln  \left( 1+3\,x \right) \right) =\\= \frac{g(x)}{(1+x)(1+2x)(1+3x)}.
$$
I want to show that $g(x)>0$. $g(0)=0$, and
$$g'(x)=\left( 1+x \right) \ln  \left( 1+x \right) -2\left( 1+2\,x \right) 
\ln  \left( 1+2\,x \right) + \left( 1+3\,x \right) \ln  \left( 1+3\,x
 \right) = h(x).
$$
Now you can guess about next step:) $h(0)=0$,
$$h'(x)=\ln  \left( 1+x \right) -4\,\ln  \left( 1+2\,x \right) +3\,\ln 
 \left( 1+3\,x \right)  = u(x)
$$
$u(0)=0$,
$$u'(x) = {\frac {2}{ \left( 1+x \right)  \left( 1+2\,x \right)  \left( 1+3\,
x \right) }} > 0.
$$
It's all.
</s>