Let $\mathrm a,b$ are positive real numbers such that for $\mathrm a - b = 10$, then the smallest value of the constant $\mathrm k$ for which $\mathrm {\sqrt {x^2 + ax}} - {\sqrt{x^2 + bx}} < k$ for all $\mathrm x>0$, is?

I don't get how to approach this problem. Any help would be appreciated.


An elementary way ($x>0$): $$f(x) = \sqrt{x^2 + ax} - \sqrt{x^2+bx} = \frac{x^2 + ax - (x^2+bx)}{\sqrt{x^2 + ax} + \sqrt{x^2+bx}} =\frac{(a - b)x}{x\sqrt{1 + \frac{a}{x}} + x\sqrt{1 + \frac{b}{x}}}= \frac{(a - b)}{\sqrt{1 + \frac{a}{x}} + \sqrt{1 + \frac{b}{x}}} < \frac{a-b}{2}= 5$$


Let $f :\mathbb{R}^+ \to \mathbb{R} \quad $such that$\quad f(x) = \sqrt{x^2 + ax} - \sqrt{x^2+bx}$

Notice that $f(x)$ is increasing when $a>b$ , $f(x) = 0$ when $a=b$ and $f(x)$ is decreasing when $a<b$.

Now when $a>b$ $\lim_{x \to \infty} f(x) = 5$

Hence $\sqrt{x^2 + ax} - \sqrt{x^2+bx} < 5 \qquad \forall a,b \in \mathbb{R}^+$

  • $\begingroup$ A question, the previous answer you posted, was there any mistake in it? Because, we didn't learn limits yet and that solution seemed like what I could understand. Just curious, thanks! $\endgroup$ – Ice Inkberry May 2 '18 at 5:32

First, replace $a$ in your inequality by $b+10$. Now the left hand side has only one parameter. Take the derivative with respect to $x$, set to zero, and find the $x$ value of the maximum as a function of $b$. Plug that in and find the left hand side maximum as a function of $b$. Take the derivative, set to zero....

  • $\begingroup$ Hello @Rossmillikan, thanks for answering, but I had this question on my test which means it had to be solved in little time. The method you proposed seems to take a lot more. I suppose there could be an easier alternative? $\endgroup$ – Ice Inkberry May 2 '18 at 2:48
  • $\begingroup$ I don't know any easier way to solve this. How much time are you given? I don't think this is so long. $\endgroup$ – Ross Millikan May 2 '18 at 2:55
  • $\begingroup$ We are given 3-4 minutes for each question $\endgroup$ – Ice Inkberry May 2 '18 at 2:56
  • $\begingroup$ I would do this in that much time. I was quicker than most, so I am probably not the best judge, but I think this the intended route to a solution $\endgroup$ – Ross Millikan May 2 '18 at 3:01
  • $\begingroup$ Umm, alright, thanks! :) $\endgroup$ – Ice Inkberry May 2 '18 at 3:04

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