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May 16th, 2016, 08:07 AM   #1
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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:

1. $\qquad$$2\,x{c}^{x-1}\ln \left( c \right)$
2. $\qquad$$ -{2}^{x}\ln \left( 2 \right)$
3. $\qquad$${c}^{x}\ln \left( c \right)$
4. $\qquad$${c}^{x}{2}^{x}\ln \left( 2 \right)$
5. $\qquad$$-x{2}^{x}\ln \left( 2 \right)$
6. $\qquad$$2\,{x}^{2}{c}^{x-1}\ln \left( c \right)$
7. $\qquad$$-{c}^{x}\ln \left( c \right) {x}^{2}$
8. $\qquad$$-{c}^{x}\ln \left( c \right) {2}^{x}$
9. $\qquad$$2\,{c}^{x-1}$
10. $\qquad$${2}^{x}$
11. $\qquad$$-2\,{c}^{x}$
12. $\qquad$${c}^{x}\ln \left( c \right) x{2}^{x}$
13. $\qquad$$-2\,x{c}^{x-1}\ln \left( c \right) {2}^{x}$
14. $\qquad$$-{c}^{x}x{2}^{x}\ln \left(2 \right)$
15. $\qquad$$2\,x{c}^{x-1}{2}^{x}\ln \left( 2 \right)$
16. $\qquad$${c}^{x}{2}^{x}$
17. $\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?
Jinliang is offline  
 
May 22nd, 2016, 02:33 AM   #2
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Joined: Aug 2012

Posts: 229
Thanks: 3

Hey Jinliang.

Have you tried finding the minimum of the function and showing it is greater than zero?

Basically you should note that you can do this numerically if the algebraic solution is difficult or too time consuming.
chiro is offline  
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