Prove (or check) the expression is positive given constraints on variables? The following proof problem have taken me a few days. Perhaps it is too hard for me to overcome it. Can you help me?
The expression is by the following:
\begin{equation}
\begin{split}
&2\,x{c}^{x-1}\ln  \left( c \right) -{2}^{x}\ln  \left( 2 \right) +{c}^
{x}\ln  \left( c \right) +{c}^{x}{2}^{x}\ln  \left( 2 \right) -x{2}^{x
}\ln  \left( 2 \right) +2\,{x}^{2}{c}^{x-1}\ln  \left( c \right) -{c}^
{x}\ln  \left( c \right) {x}^{2}\\
&-{c}^{x}\ln  \left( c \right) {2}^{x}+
2\,{c}^{x-1}+{2}^{x}-2\,{c}^{x}+{c}^{x}\ln  \left( c \right) x{2}^{x}-
2\,x{c}^{x-1}\ln  \left( c \right) {2}^{x}-{c}^{x}x{2}^{x}\ln  \left( 
2 \right) \\
&+2\,x{c}^{x-1}{2}^{x}\ln  \left( 2 \right) +{c}^{x}{2}^{x}-2
\,{c}^{x-1}{2}^{x},
\end{split}
\end{equation}
where $x\in[2,+\infty)$, and $1<c<2$.
Our goal is to prove that the aforementioned expression is positive .
To facilitate subsequent view, I give each terms in the expression a unique sequence number by the following:


*

*$\qquad$$2\,x{c}^{x-1}\ln  \left( c \right)$

*$\qquad$$ -{2}^{x}\ln  \left( 2 \right)$

*$\qquad$${c}^{x}\ln  \left( c \right)$

*$\qquad$${c}^{x}{2}^{x}\ln  \left( 2 \right)$

*$\qquad$$-x{2}^{x}\ln  \left( 2 \right)$

*$\qquad$$2\,{x}^{2}{c}^{x-1}\ln  \left( c \right)$

*$\qquad$$-{c}^{x}\ln  \left( c \right) {x}^{2}$

*$\qquad$$-{c}^{x}\ln  \left( c \right) {2}^{x}$

*$\qquad$$2\,{c}^{x-1}$

*$\qquad$${2}^{x}$

*$\qquad$$-2\,{c}^{x}$

*$\qquad$${c}^{x}\ln  \left( c \right) x{2}^{x}$

*$\qquad$$-2\,x{c}^{x-1}\ln  \left( c \right) {2}^{x}$

*$\qquad$$-{c}^{x}x{2}^{x}\ln  \left(2 \right)$

*$\qquad$$2\,x{c}^{x-1}{2}^{x}\ln  \left( 2 \right)$

*$\qquad$${c}^{x}{2}^{x}$

*$\qquad$$-2\,{c}^{x-1}{2}^{x}$


Maybe the right way is $\cdots\quad$Try showing that each term  is $>0$. If some are $< 0$, try combining two or more. This will get you closer to the desired proof.
But HOW?

The following is a false proof. However, the method may be right. So it still has certain reference significance.
Combine term $9$, $10$ and $11$, we get
$$
2\,{c}^{x-1}+{2}^{x}-2\,{c}^{x}=\left( \frac{2}{c}-1 \right) {c}^{x}+\left({2}^{x}-{c}^{x}\right)>0
$$;
Combine term $4$ and $8$, we get
$$
{c}^{x}{2}^{x}\ln  \left( 2 \right) -{c}^{x}\ln  \left( c \right) {2}^{x}=c^x2^x\left(\ln(2)-\ln(c)\right)>0
$$;
Combine term $(12,14) + (13,15) $, we get
\begin{equation}
\begin{split}
\,&\{{c}^{x}\ln  \left( c \right) x{2}^{x}-{c}^{x}x{2}^{x}\ln  \left(2 \right)\}+\{-2\,x{c}^{x-1}\ln  \left( c \right) {2}^{x}+2\,x{c}^{x-1}{2}^{x}\ln  \left( 2 \right)\}\\
&=-c^xx2^x\left(\ln2-\ln c\right)+2xc^{x-1}2^x\left(\ln2-\ln c\right)\\
&=\left(\frac{2}{c}-1\right)2^xxc^x\left(\ln2-\ln c\right)>0
\end{split}
\end{equation}
;
Combine term $1,3,6,7$, we get
\begin{equation}
\begin{split}
&2\,x{c}^{x-1}\ln  \left( c \right)+{c}^{x}\ln  \left( c \right)+2\,{x}^{2}{c}^{x-1}\ln  \left( c \right)-{c}^{x}\ln  \left( c \right) {x}^{2}\\
&=\left(\left(\frac{2}{c}-1\right)x^2+\frac{2}{c}x+1\right)\cdot c^x\ln c
\end{split}.
\end{equation}
As the equation
$$\left(\frac{2}{c}-1\right)x^2+\frac{2}{c}x+1=0$$ has two roots $-1$ and $-\frac{1}{\frac{2}{c}-1}$, which has the following inequality relation
$$-\frac{1}{\frac{2}{c}-1}<-1,$$
So we have $$\left(\frac{2}{c}-1\right)x^2+\frac{2}{c}x+1>0$$ 
for $1<c<2$ and $x>2$
.
At this point, four terms have not been considered, namely $ 2,5,16$ and $17$. One can easily deduce that 2+5 and 16+17 are both negative quantities. 
\begin{equation}
\begin{split}
&\{-{2}^{x}\ln  \left( 2 \right) -x{2}^{x}\ln  \left( 2 \right)\} +\{{c}^{x}{
2}^{x}-2\,{c}^{x-1}{2}^{x}\}\\
&=(-1-x)2^x\ln 2 +(1-\frac{2}{c})c^x2^x<0
\end{split}
\end{equation}
So something went wrong.
