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show that

$$\dfrac{2^x-1}{3^x-2^x}\le3\left(\dfrac{1}{x-1}-\dfrac{1}{x}\right)\cdots(1)$$

This problem (1) is from when I solve following $$1+\dfrac{3}{5}+\dfrac{7}{19}+\cdots+\dfrac{2^n-1}{3^n-2^n}\le 4-\dfrac{3}{n}\cdots (2)$$

My try if we have prove this $$\dfrac{2^n-1}{3^n-2^n}\le 3(\dfrac{1}{n-1}-\dfrac{1}{n}),n\ge 2$$ then $(2)$ is prove it.

By the way,I have use other methods solve it $(2)$

because I have try

$$\dfrac{2^n-1}{3^n-2^n}\le\left(\dfrac{2}{3}\right)^{n-1}$$ $$\Longleftrightarrow 6^n+2\cdot 3^n\ge 3\cdot 4^n$$ since $AM-GM$ $$\dfrac{1}{3}\cdot 6^n+\dfrac{2}{3}\cdot 3^n\ge\left(\dfrac{1}{3}\cdot 6+\dfrac{2}{3}\cdot 3\right)^n=4^n$$ so $$\sum_{k=1}^{n}\dfrac{2^k-1}{3^k-2^k}\le\sum_{k=1}^{n}\left(\dfrac{2}{3}\right)^{k-1}=3-3\cdot\left(\dfrac{2}{3}\right)^n$$ and it easy prove $$4-\dfrac{3}{n}\ge 3-3\cdot(\dfrac{2}{3})^n$$ so the $(2)$ prove by done.

Now my question How can prove by $(1)$,and I think $(1)$ have lot of nice methods.Thank you

Andrés E. Caicedo
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  • Dupe of http://math.stackexchange.com/questions/504900/how-prove-this-equation-all-real-roots but that has no answers yet. – Daniel Fischer Sep 25 '13 at 17:40
  • No,I think this equality is true.see,http://www.wolframalpha.com/input/?i=%282%5Ex-1%29%2F%283%5Ex-2%5Ex%29-3%281%2F%28x-1%29-1%2Fx%29,But Now I can't prove it.and this is nice problem. – math110 Sep 25 '13 at 17:51
  • When $x=0.1$ the inequality is the reverse. Some restriction on $x$ is required, maybe to make the right side positive. – coffeemath Sep 25 '13 at 18:08
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    @coffeemath The restricted that they stated (buried in the question) is $x \geq 2$. – Calvin Lin Sep 25 '13 at 18:26
  • Yes,I think $x> 1$.is true.@CalvinLin,Have you nice methods? – math110 Sep 26 '13 at 02:36

1 Answers1

1

We can show the stronger inequality (i.e. with increased left side) $$\dfrac{2^x}{3^x-2^x}\le3\left(\dfrac{1}{x-1}-\dfrac{1}{x}\right)\tag{1} $$ holds for $x\ge 6.$ Division of the numerator and denominator on the left by $2^x$ and algebra gives the equivalent form $$(3/2)^x \ge \frac{x^2-x+3}{3}. \tag{2}$$ This holds when $x=6$, and one can show the derivative of the left side exceeds that of the right side for $x\ge 6.$ If we increase the right side derivative to $2x/3$, then one could get exact values [using Lambert's W function] for when $\ln(3/2)\cdot(3/2)^x$ crosses through the line $y=2x/3$, and these are approximately at $0.863$ and $5.373$, so that for $x\ge 6$ we have the derivative of the left side of $(2)$ exceeds that of the right side, while also $(2)$ holds at $x=6$, from which we can conclude $(2)$, and so also $(1)$, hold for $x\ge 6$

Note: the inequality $(1)$ fails to hold only on a short interval $(\alpha,\beta)$ contained in the interval $[4,6]$. The above approach doesn't fill that gap, and it seems (to me) that without dropping the subtraction of $1$ in the given inequality makes the manipulations more complicated, making an approach via derivatives difficult.

coffeemath
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