If a,b are positive real numbers such that $a-b=2$, then find the smallest value of the constant L for which $\sqrt{x^2+ax}-\sqrt{x^2+bx}<L$ $\forall$ $x>0$
$$y=\sqrt{x^2+ax}-\sqrt{x^2+bx}$$ $$y=\sqrt{x^2+ax}-\sqrt{x^2+ax-2x}$$
Let $x^2+ax=t$
Squaring both sides $$y^2=\sqrt{t}-\sqrt{t-2x}$$ $$y^2=t+t-2x-2\sqrt{t}\sqrt{t-2x}$$ $$y^2=2t-2x-2\sqrt{t^2-2xt}$$ $$y^2-2t+2x=-2\sqrt{t^2-2xt}$$
Squaring both sides again $$y^4+4t^2-4y^2t+4x^2+4x(y^2-2t)=4(t^2-2xt)$$ $$y^4+4t^2-4y^2t+4x^2+4xy^2-8xt=4t^2-8xt$$ $$4x^2+4xy^2+y^4-4y^2t=0$$
$$4x^2+4xy^2+y^4-4y^2(x^2+ax)=0$$ $$4x^2(1-y^2)+4xy^2(1-a)+y^4=0$$ $$x=\dfrac{-4y^2(1-a)\pm\sqrt{16y^4(1-a)^2-16y^4(1-y^2)}}{8(1-y^2)}$$ $$x=\dfrac{-4y^2(1-a)\pm4y^2\sqrt{1+a^2-2a-1+y^2}}{8(1-y^2)}$$ $$x=\dfrac{-4y^2(1-a)\pm4y^2\sqrt{a^2-2a+y^2}}{8(1-y^2)}$$ $$x=\dfrac{-y^2(1-a)\pm y^2\sqrt{a^2-2a+y^2}}{2(1-y^2)}$$
We want to find the range of $y$ for $x>0$
Let's first take $x$ with positive sign
$$\dfrac{-y^2(1-a)+ y^2\sqrt{a^2-2a+y^2}}{2(1-y^2)}>0$$ $$\dfrac{(a-1)+\sqrt{a^2-2a+y^2}}{1-y^2}>0$$
As $a=b+2$, hence $a>1$ as $a,b$ are positive numbers
Hence numerator is always positive, so
$$1-y^2>0$$ $$y^2-1<0$$ $$y\in(-1,1)$$
But as $y=\sqrt{x^2+ax}-\sqrt{x^2+bx}$, therefore $y>0$ as $a>b$
So $y\in(0,1)$
Let's take $x$ with negative sign
$$\dfrac{-y^2(1-a)- y^2\sqrt{a^2-2a+y^2}}{2(1-y^2)}>0$$ $$\dfrac{(a-1)- \sqrt{a^2-2a+y^2}}{2(1-y^2)}>0$$ $$\dfrac{(a-1)- \sqrt{(a-1)^2+y^2-1}}{2(1-y^2)}>0$$
If $y^2-1>0$, then numerator will be negative, hence denominator should be negative, it means $1-y^2<0$ which is contradictory.
If $y^2-1<0$, then numerator will be positive, hence denominator should be positive, it means $y^2-1>0$ which is contradictory.
So only previous inequality is valid, hence $y\in(0,1)$
Thus value of L is $1$
But this solution is very long, is there any shorter way to do this? Please help me in this.
Just write
$$\sqrt{x^2+ax}-\sqrt{x^2+bx} = \frac{(a-b)x}{\sqrt{x^2+ax}+\sqrt{x^2+bx}} \stackrel{x>0}{=} \frac{a-b}{\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}} < \frac{2}{2} = 1$$
And note that $\frac{2}{\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}}$ is strictly increasing with $\lim_{x\to \infty}\frac{2}{\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}} = 1$
So, $\boxed{L=1}$.