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Is there a way to show that this expression is always positive as long as $b>0$ and $r>x$ ? Assume $r>0$ and $x>0$ \begin{equation} b-1+\sqrt{(1+b)^2-\frac{4rb}{x}} \end{equation}
It's simple with $b>1$ but I can't figure it out if it is true for lower values.
The statement is obviously not true if $(1+b)^2 - \frac {4br}x < 0$.
Case 1: $(1+b)^2 - \frac {4br}x < 0$ or in other words
$(1+b)^2 < \frac {4br}x$ or
$\frac {(1+b)^2*|x|}{4|b|} < r$. (Assuming $b \ne 0$. If $b = 0$ then $(1+b)^2 - \frac {4br}x = (1+b)^2 = 1$.)
The statement is false and yields non-real number.
Case 2:
$(1+b)^2 - \frac {4br}x = 0$ or $r =\frac {(1+b)^2*|x|}{4|b|}$
In this case $b-1 + \sqrt{(1+b)^2 - \frac {4br}x} = b-1 > 0 \iff b > 1$
Case 3:$r <\frac {(1+b)^2*|x|}{4|b|}$
$b-1 + \sqrt{(1+b)^2 - \frac {4br}x} > 0 \iff$
$\sqrt{(1+b)^2 - \frac {4br}x} > 1-b$.
Case 3a: This will be true for $b \ge 1$.
If $b < 1$ then this will be true if
$(1+b)^2 - \frac {4br}x > (1-b)^2 \iff$
$1 + b^2 + 2b - \frac {4br}x > 1 + b^2 -2b \iff$
$4b > \frac {4br}x \iff \frac b{|b|} > \frac rx$
Case 3b: $1> b > 0$ then true if $|x| > r$
Case 3c: $-1< b < 0$ then this is true if $0 > -|x| >r$
So given $b > 0$ and $r > x > 0$ we must have $b \ge 1$ and $x< r \le\frac {(1+b)^2x}{4b}$ or in other words $1 < \frac rx \le \frac{(1+b)^2}{4b}$.
But if $b = 1$ we have $1 < \frac rx \le 1$ so $b > 1$. But we must also have the condition $x< r \le\frac {(1+b)^2x}{4b}$.
This is false. Consider $b = \frac{1}{10}, x = 1, r = \frac32$. Clearly, we have $b,x,r>0$ and $r>x$.
$$\sqrt{(1+b)^2-\frac{4rb}{x}} = \sqrt{\left(1+\frac{1}{10}\right)^2-\frac{4(3/2)(1/10)}{1}} = \sqrt\frac{121-60}{100} = \frac{\sqrt{61}}{10}$$
$$b-1+\sqrt{(1+b)^2-\frac{4rb}{x}} = -\frac9{10} + \frac{\sqrt{61}}{10} = \frac{\sqrt{61}-\sqrt{81}}{10} < 0$$
