Let $\mathrm a,b$ are positive real numbers such that for $\mathrm a - b = 10$, then the smallest value of the constant $\mathrm k$ for which $\mathrm {\sqrt {x^2 + ax}} - {\sqrt{x^2 + bx}} < k$ for all $\mathrm x>0$, is?
I don't get how to approach this problem. Any help would be appreciated.
On
First, replace $a$ in your inequality by $b+10$. Now the left hand side has only one parameter. Take the derivative with respect to $x$, set to zero, and find the $x$ value of the maximum as a function of $b$. Plug that in and find the left hand side maximum as a function of $b$. Take the derivative, set to zero....
On
Let $f :\mathbb{R}^+ \to \mathbb{R} \quad $such that$\quad f(x) = \sqrt{x^2 + ax} - \sqrt{x^2+bx}$
Notice that $f(x)$ is increasing when $a>b$ , $f(x) = 0$ when $a=b$ and $f(x)$ is decreasing when $a<b$.
Now when $a>b$ $\lim_{x \to \infty} f(x) = 5$
Hence $\sqrt{x^2 + ax} - \sqrt{x^2+bx} < 5 \qquad \forall a,b \in \mathbb{R}^+$
An elementary way ($x>0$): $$f(x) = \sqrt{x^2 + ax} - \sqrt{x^2+bx} = \frac{x^2 + ax - (x^2+bx)}{\sqrt{x^2 + ax} + \sqrt{x^2+bx}} =\frac{(a - b)x}{x\sqrt{1 + \frac{a}{x}} + x\sqrt{1 + \frac{b}{x}}}= \frac{(a - b)}{\sqrt{1 + \frac{a}{x}} + \sqrt{1 + \frac{b}{x}}} < \frac{a-b}{2}= 5$$