# How to prove this inequality $\Big| \frac{-a + \sqrt{a^2-b^2}}{b} \Big| < 1$

Suppose $$a > b > 0$$,

how to prove that $$\Big| \frac{-a + \sqrt{a^2-b^2}}{b} \Big| < 1$$

I been working on this for like 2 hours still did not find the trick.

Since when $$a = 5$$ and $$b = 4$$, this inequality holds, but what's the trick to actually prove it?

Note that$$\left|-a+\sqrt{a^2-b^2}\right|=a-\sqrt{a^2-b^2},$$since $$0, and that\begin{align}a-\sqrt{a^2-b^2}which is true, since $$b>0$$.

Note that $$0 < a^2 - b^2 < a^2$$. So $$\sqrt{a^2-b^2} < a$$. Hence

\begin{align} \left| \frac{-a + \sqrt{a^2-b^2}}{b} \right| &= \frac{a - \sqrt{a^2-b^2}}{b} \\ &= \dfrac{b}{a+\sqrt{a^2-b^2}} \\ &< 1 \end{align}

The number in question is the absolute value of the greatest root of $$\frac{1}{2}b x^2+ax+\frac{1}{2}b$$. Let $$\alpha$$ denote what is inside the absolute value. Then $$\alpha$$ must be negative since $$a^2-b^2 < a^2$$. Meanwhile, by symmetry, if $$x_{0}$$ is a non-zero root of this polynomial, then so is $$1/x_{0}$$. Since $$\alpha$$ is the greatest root and negative, we must have $$1/\alpha < - 1 < \alpha <0$$, and the result follows.

Since $$a>b$$, put $$x=a/b >1$$. Then the inequality you are trying to prove becomes $$\left|-x + \sqrt{x^2-1}\right| < 1.$$

Now this is quite straight forward.

Infact, we can see $$\left|(-x + \sqrt{x^2-1})(x + \sqrt{x^2-1})\right| = 1,$$

and therefore

$$\left|-x + \sqrt{x^2-1}\right| < \frac{1}{(x + \sqrt{x^2-1})}.$$

The quadratic equation $$bx^2+2ax+b=0$$ has two roots:

$$r=\frac{-a+\sqrt{a^2-b^2}}{b}$$ and $$s=-\frac{a+\sqrt{a^2-b^2}}{b}$$.

Obviously, $$|s|>1$$, so $$|r|=\frac{1}{|s|}<1$$ from Vieta's formulas.

WLOG $$b=a\sin2t,$$

As $$a>b>0,0<\sin2t<1$$

WLOG $$0<2t<\dfrac\pi2\implies 0

$$\dfrac{-a+\sqrt{a^2-b^2}}b=\dfrac{-a+a\cos2t}{a\sin2t}=\dfrac{1-2\sin^2t-1}{2\sin t\cos t}=-\tan t$$

By $$(1), -1<\tan t<0$$