Truth about derivative of a function

We can write $\frac{1}{r(a)} \; = \; \frac{1}{h} + \frac{1}{a} + \sqrt{\frac{1}{h^2} + \frac{1}{a^2}}$ which is clearly a strictly decreasing function of $a$ , and hence $r$ is a strictly increasing function of $a$ .

The function simplifies to:

$r(a) = \cfrac{ha}{a + h + \sqrt{a^2 + h^2}} = \cfrac{1}{2}(a + h - \sqrt{a^2 + h^2})$

and its derivative is:

$r'(a) = \cfrac{1}{2}\bigg(1 - \cfrac{a}{\sqrt{a^2 + h^2}}\bigg)$

Now $a^2 + h^2 > a^2$ , so $1 > \cfrac{a^2}{a^2 + h^2}$ , and $1 > \cfrac{a}{\sqrt{a^2 + h^2}}$ , and $1 - \cfrac{a}{\sqrt{a^2 + h^2}} > 0$ , so $\cfrac{1}{2}\bigg(1 - \cfrac{a}{\sqrt{a^2 + h^2}}\bigg) > 0$ . Therefore,

$r'(a) = \cfrac{1}{2}\bigg(1 - \cfrac{a}{\sqrt{a^2 + h^2}}\bigg) > 0$

which means $r(a)$ is a strictly increasing function of $a$ .

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(The function is also the equation of the radius $r$ of the incircle of a right triangle with legs $a$ and $h$ . Therefore, as $a$ increases, $r$ increases.)

The function $r(a)$ will be strictly increasing iff $r'(a) > 0$ for all $a, h > 0.$ This derivative computes to:

$r'(a) = \frac{h(a+h+\sqrt{a^2+h^2}) - (ha)(1 + a/ \sqrt{a^2+h^2})}{(a+h+\sqrt{a^2+h^2})^2} = \boxed{\frac{h^2 \sqrt{a^2+h^2} + h^3}{(a+h+\sqrt{a^2+h^2})^{5/2}}}$

which is positive for all $a,h>0 \Rightarrow r(a)$ is a strictly increasing function.

If we apply the quotient rule, the derivative w.r.t. $a$ has the same sign as

$h(a+h+\sqrt{a^2+h^2})-ha(1+0+\frac{a}{\sqrt{a^2+h^2}})=h(h+\frac{a^2+h^2-a^2}{\sqrt{a^2+h^2}})>0$