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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?

FlyFish
  • 89

1 Answers1

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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.

FlyFish
  • 89