# Proof that expression is positive if conditions is met

Is there a way to show that this expression is always positive as long as $b>0$ and $r>x$ ? Assume $r>0$ and $x>0$ $$b-1+\sqrt{(1+b)^2-\frac{4rb}{x}}$$

It's simple with $b>1$ but I can't figure it out if it is true for lower values.

• Just prove $\sqrt{(1+b)^2-\frac{4rb}{x}} \gt 1- b$
– user261263
Feb 25, 2017 at 17:54
• If $b > 0$ then $(1+b)^2 > 1$. if $r>x$... hmm. Are we told $x > 0$ ? If so $r/x > 0$ and $(1+ b)^2 - 4xb/x > (1+b)^2 - 4b = 1 - 2b + b^2 = (1-b)^2$. Can you go from there. Feb 25, 2017 at 17:58
• The expression inside the square root can be made negative by taking $r$ extremely large. Feb 25, 2017 at 18:07

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 $x< r \le\frac {(1+b)^2x}{4b}$.

• Have you seen my counterexample? Even though we have $stuff \ge 0$ and $0 < b \le 1$, the inequality is still false. Feb 25, 2017 at 18:12
• Yeah.. Not exactly seeing were my error is. $1.1^2 - .6 > 1.1^2 - .4$ ... Oh, god! That was stupid!!!! Feb 25, 2017 at 18:25

This is false. Consider $b = \frac{1}{10}, x = 1, r = \frac32$. Clearly, we have $b,x,r>0$ and $r>x$.

$$\sqrt{(1+b)^2-\frac{4rb}{x}} = \sqrt{\left(1+\frac{1}{10}\right)^2-\frac{4(3/2)(1/10)}{1}} = \sqrt\frac{121-60}{100} = \frac{\sqrt{61}}{10}$$

$$b-1+\sqrt{(1+b)^2-\frac{4rb}{x}} = -\frac9{10} + \frac{\sqrt{61}}{10} = \frac{\sqrt{61}-\sqrt{81}}{10} < 0$$

• Oh, nice counter-example! Thanks
– Dio
Feb 25, 2017 at 18:06