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

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