<s>[INST] Let $a, b \in \mathbb{R^+}$ such that $a-b=10$, find smallest value of constant $K$ for which $\sqrt{(x^2 + ax)}- \sqrt {(x^2 +bx)}0$. 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
 [/INST] 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 $x$
$$
\sqrt{\frac{a}{x}+1}-\sqrt{\frac{b}{x}+1} = \frac{a-b}{2 x}+\frac{1}{8} \left(\frac{1}{x}\right)^2
   \left(b^2-a^2\right)+O\left(\left(\frac{1}{x}\right)^3\right)
$$
</s>