Prove of an inequality between the radius and segments of an incircle Let the triangle $ABC$ have an incircle with tangent points $G,E,F$ with $G \in$ $\overline {AC}$, $E \in \overline {AB}$ and $F \in \overline {BC}$.
If $\frac {GC}{AG}= p$ and $\frac {FB}{CF}= q$ and $p+q=1$. Prove that the radius $r$ holds the next inequality:
$$ {r\le \frac {AE}{2 \sqrt 2} }$$
Things i saw
I saw that $AG = AE$, $EB = FB$ and $CG=CF$ and from another problem that i did, if $p+q=1$ then $pq \le \frac{1}{4}$, so $\frac {GC}{AG}$ $\cdot$ $\frac {FB}{CF} \le \frac{1}{4}$ but i don't know if the last inequality helps for the principal one

 A: Assume that $\overline{AE} = \overline{AG} = u$, $\overline{CG} = \overline{CF} = v$, $\overline{BE} = \overline{BF} = w$. By Heron's formula, the area of $\triangle ABC = \sqrt{(u+v+w)uvw}$, and $r(u+v+w) = \triangle ABC$, hence
$$
r = \sqrt{\frac{uvw}{u+v+w}}, \mbox{ and } \frac ru = \sqrt{\frac{vw}{u(u+v+w)}}.
$$
So it suffices to show
$$
\frac{u(u+v+w)}{vw} \geq 8,
$$
with the constraint $\frac vu + \frac wv = 1$ (note that since $u, v, w > 0$ this implicitly implies $u > v > w$). Since this is homogeneous we may assume $w = 1$, then you get $u = \frac{v^2}{v-1}$, and this reduces to 
$$
\frac{v(2v^2 - 1)}{(v-1)^2} = 2v + 4 + \frac{5v-4}{(v-1)^2} \geq 8, \quad \mbox{where $v > 1$}.
$$
It is not hard to show that this is true. Actually the minima of $\frac{u(u+v+w)}{vw}$ is around 12.6, according to WolframAlpha.
A: Let $AG=x$, $FB=y$, $GC=z$ and $y=tz$.
Hence, the condition gives $$\frac{z}{x}+\frac{y}{z}=1$$ or
$$x=\frac{z}{1-t},$$ where $0<t<1$ and we need to prove that
$$\frac{S_{\Delta ABC}}{x+y+z}\leq\frac{x}{2\sqrt2}$$ or
$$\frac{\sqrt{(x+y+z)xyz}}{x+y+z}\leq\frac{x}{2\sqrt2}$$ or
$$8yz\leq x(x+y+z)$$ or
$$8t\leq\frac{1}{1-t}\left(\frac{1}{1-t}+t+1\right)$$ or
$$8t^3-15t^2+8t-2\leq0,$$ which is true because
$$8t^3-15t^2+8t-2<9t^3-15t^2+8t-2=(t-1)(9t^2-6t+2)<0.$$
