# Least value of $L$ for $\sqrt{x^2+ax}-\sqrt{x^2+bx}<L$ for all $x>0$

If $$a$$ and $$b$$ are positive real numbers such that $$a-b=2$$ , then the smallest value of the constant $$L$$ for which $$\sqrt{x^2+ax}-\sqrt{x^2+bx} for all $$x>0$$.

This question is similar to this one, but I don't want to apply the concept of limit. I want to use application of derivative in solving this problem.

I thought of putting $$\sqrt{x^2+ax}-\sqrt{x^2+bx}=L$$ and then using $$\frac{dL}{dx}=0$$, but the equation is coming in the form of $$a+b$$ which I am not able to solve.

We have $$\sqrt{x^2+ax}-\sqrt{x^2+bx}=\frac{(x^2+ax)-(x^2+bx)}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}=\frac{a-b}{\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}}$$ for all $$x>0$$. Because $$\sqrt{1+\frac{a}{x}}>1$$ and $$\sqrt{1+\frac{b}{x}}>1$$ for all $$x>0$$, we get $$\sqrt{x^2+ax}-\sqrt{x^2+bx}< \frac{a-b}{2}.$$ This shows that $$L\leq \frac{a-b}{2}$$. As $$\lim_{x\to\infty}\left(\sqrt{x^2+ax}-\sqrt{x^2+bx}\right)=\lim_{x\to\infty}\frac{a-b}{\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}}=\frac{a-b}{2},$$ we must have $$L=\frac{a-b}{2}$$. (By the way, you cannot avoid using limits in this problem because $$L$$ is attained as $$x$$ goes to infinity. Taking derivative of $$\sqrt{x^2+ax}-\sqrt{x^2+bx}$$ will not give you a maximizing point on $$(0,\infty)$$ because the maximum does not exist.)