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For all integers $x\geq 7$ and all real numbers $y$ such that $0.249 \leq y \leq 0.25$ define $$F(x,y)=-\frac{\log\left( \frac{x+3}{x}\right)+\frac{1}{2}\log\left(\frac{x^2}{x^2-1} \right)+\frac{1-4y}{2}\log\left( \frac{x+1}{x-1} \right)-2\sqrt{\frac{x+3}{x}}+3-4y}{-\log\left( \frac{x+3}{x}\right)+2\sqrt{\frac{x+3}{x}}+\log\left( \frac{x+1}{x}\right)-2\sqrt{\frac{x+1}{x}}}.$$

I claim that $$F(x,y)\geq F(7,y)$$ for all $x \geq 7$.

I thought of several ways to try and prove this:

  1. Working directly on this inequality is complicated (even Maple couldn't do it for me).

  2. I plotted $F(x,y)$ for fixed values of $y$ and saw that $F(x,y)$ was increasing with respect to $x$ in all of the plots. However, the derivative of $F(x,y)$ with respect to $x$ is very complicated as well.

  3. Since the values of $x$ are integers, I thought about an inductive argument, like showing that $F(x,y)\leq F(x+1,y)$ for all $x\geq 7$, but again, this is a difficult inequality to work with.

***If it is too difficult to show that $F(x,y)\geq F(7,y)$, then I would also be very happy to figure out how to at least show that $$F(x,y)\geq 1.05-y$$ for all $x\geq 7$ and $y$ sufficiently close to $0.25$.

Any suggestions are greatly appreciated. Thank you.

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    $\begingroup$ Could there be a typo in the expression? Let $y = 0.25-\epsilon$ for some small $\epsilon > 0$ and take a limit as $x \rightarrow \infty$. Assuming I didn't make any stupid mistakes, the numerator converges to $4\epsilon$ and the denominator converges to $0+$. Thus, $\lim_{x \to \infty} F(x,y) = -\infty < 0.8+\epsilon = 1.05-y$. $\endgroup$ Feb 6, 2021 at 0:35

1 Answer 1

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$F(x,y)=-\frac{\log\left( \frac{x+3}{x}\right)+\frac{1}{2}\log\left(\frac{x^2}{x^2-1} \right)+\frac{1-4y}{2}\log\left( \frac{x+1}{x-1} \right)-2\sqrt{\frac{x+3}{x}}+3-4y}{-\log\left( \frac{x+3}{x}\right)+2\sqrt{\frac{x+3}{x}}+\log\left( \frac{x+1}{x}\right)-2\sqrt{\frac{x+1}{x}}}$
$=-\frac{\log\left( \frac{x+3}{x}\right)+\frac{1}{2}\log\left(\frac{x^2}{x^2-1} \right)+\frac{1-4y}{2}\log\left( \frac{x+1}{x-1} \right)-2\sqrt{\frac{x+3}{x}}+3-4y}{-\log\left( x+3\right)+2\sqrt{\frac{x+3}{x}}+\log\left( x+1\right)-2\sqrt{\frac{x+1}{x}}}$
$=-\frac{\log\left( x+3\right)-\frac{1}{2}\log\left(x^2-1 \right)+\frac{1-4y}{2}\log\left( \frac{x+1}{x-1} \right)-2\sqrt{\frac{x+3}{x}}+3-4y}{-\log\left( x+3\right)+2\sqrt{\frac{x+3}{x}}+\log\left( x+1\right)-2\sqrt{\frac{x+1}{x}}}$
$=-\frac{\log\left( \frac{x+3}{\sqrt{x^2-1}}\right)+\frac{1-4y}{2}\log\left( \frac{x+1}{x-1} \right)-2\sqrt{\frac{x+3}{x}}+3-4y}{\log\left( \frac{x+1}{x+3}\right)+2\sqrt{\frac{x+3}{x}}-2\sqrt{\frac{x+1}{x}}}$
$=-\frac{\log\left( \frac{x+3}{\sqrt{x^2-1}}\right)+\frac{1}{2}\log\left( \frac{x+1}{x-1} \right)-y\log\left( \frac{x+1}{x-1} \right)-2\sqrt{\frac{x+3}{x}}+3-4y}{\log\left( \frac{x+1}{x+3}\right)+2\sqrt{\frac{x+3}{x}}-2\sqrt{\frac{x+1}{x}}}$
$=-\frac{\log\left( \frac{x+3}{\sqrt{x^2-1}}\frac{\sqrt{x+1}}{\sqrt{x-1}}\right)-y\log\left( \frac{x+1}{x-1} \right)-2\sqrt{\frac{x+3}{x}}+3-4y}{\log\left( \frac{x+1}{x+3}\right)+2\sqrt{\frac{x+3}{x}}-2\sqrt{\frac{x+1}{x}}}$
$=-\frac{\log\left( \frac{x+3}{\sqrt{x+1}\sqrt{x-1}}\frac{\sqrt{x+1}}{\sqrt{x-1}}\right)-y\log\left( \frac{x+1}{x-1} \right)-2\sqrt{\frac{x+3}{x}}+3-4y}{\log\left( \frac{x+1}{x+3}\right)+2\sqrt{\frac{x+3}{x}}-2\sqrt{\frac{x+1}{x}}}$
$=-\frac{\log\left( \frac{x+3}{x-1}\right)-y\log\left( \frac{x+1}{x-1} \right)-2\sqrt{\frac{x+3}{x}}+3-4y}{\log\left( \frac{x+1}{x+3}\right)+2\sqrt{\frac{x+3}{x}}-2\sqrt{\frac{x+1}{x}}}$
$=\frac{\log\left( \frac{x-1}{x+3}\right)+y\log\left( \frac{x+1}{x-1} \right)+2\sqrt{\frac{x+3}{x}}-3+4y}{\log\left( \frac{x+1}{x+3}\right)+2\sqrt{\frac{x+3}{x}}-2\sqrt{\frac{x+1}{x}}}$
Now it should be easier to take the derivative or do other simplification steps.

EDIT: As mentioned in the comments, it seems to be decreasing instead of increasing. That could you also show by the last term of above simplification.

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