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!
 A: 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.
A: 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.
A: It's not true.
Try plugging in a large value for t such as $10^9$ (tried in R and sage)
A: EDIT:  I made a typo, entering the function into the graphing software, leading to the following believable conjecture (based on the incorrect graph).  I'll leave the answer up for posterity, but please don't upvote it.
Conjecture:  


*

*$f(t) > 0$ for $0 < t < 1$,

*$f(t) < 0$ for $t > 1$,

*$f(1) = 0$, and $t = 1$ is the only root,

*$f'(t) < 0$ for all $t > 0$ ($f$ is always decreasing),

*$f''(t) > 0$ for all $t > 0$ ($f$ is always concave up).




Here's a better graph, which suggests that the inequality in the question is correct.

By the way, you can have a look at the graph and manipulate the zoom here.
A: I don't think it is true. I test the expression from $t = 10^{10}+1$ to $t = 10^{10}+10000$ in Matlab. The min is about $-9.1\times 10^{-13}$. It may be due to the floating error. However I think the limit when $t$ tends to $\infty$ is 0 but it would not be constantly positive.
A: 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.
