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

Question

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?

Answer 1

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

Answer 2

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}

Answer 3

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.

Answer 4

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

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

Answer 5

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.

Answer 6

WLOG $b=a\sin2t,$

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

WLOG $0<2t<\dfrac\pi2\implies 0<t<\dfrac\pi4 \ \ \ \ (1)$

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

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