Let $a, b \in \mathbb{R^+}$ such that $a-b=10$, find smallest value of constant $K$ for which $\sqrt{(x^2 + ax)}- \sqrt {(x^2 +bx)}<K$ for all $x>0$.

Question

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

My try:

Unable to solve further

Answer 1

Hint.

$$ \sqrt{(x^2 + ax)} - \sqrt {(x^2 +bx)} = x\left(\sqrt{\frac{a}{x}+1}-\sqrt{\frac{b}{x}+1}\right) $$

and for large $x$

$$ \sqrt{\frac{a}{x}+1}-\sqrt{\frac{b}{x}+1} = \frac{a-b}{2 x}+\frac{1}{8} \left(\frac{1}{x}\right)^2 \left(b^2-a^2\right)+O\left(\left(\frac{1}{x}\right)^3\right) $$

Answer 2

Using $$ \sqrt{a}-\sqrt{b} = {a-b\over \sqrt{a}+\sqrt{b}}$$

so $${(x^2+ax)-(x^2+bx)\over \sqrt{(x^2 + ax)} + \sqrt {(x^2 +bx)}} < K$$

thus $${10x\over \sqrt{(x^2 + ax)} + \sqrt {(x^2 +bx)}} < K$$

taking $x\to \infty$ we get $$\lim _{x\to \infty}{10\over \sqrt{1 + {a\over x}} + \sqrt {1 +{b\over x}}} = 5 \leq K$$