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

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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.)

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