Let be $\{I\}=AD\cap BE$. We have $DI=\frac{2abc}{(a+b+c)(b+c)}\cos\frac A2$ and $EI=\frac{2abc}{(a+b+c)(a+c)}\cos\frac B2$. Therefore from the law of cosines in $\triangle DIE$ we have: \[DE^2=\frac{a^2b^2c^2}{s^2}\left[\frac{\cos^2\frac A2}{(b+c)^2}+\frac{\cos^2\frac B2}{(a+c)^2}+2\frac{\cos\frac A2\cos\frac B2\sin\frac C2}{(a+c)(b+c)}\right]=\]\[\frac{abc}{s}\left[\frac{a(s-a)}{(b+c)^2}+\frac{b(s-b)}{(a+c)^2}+\frac{2(s-a)(s-b)}{(a+c)(b+c)}\right]\] WLOG suppose $a=y+z, b=x+z,c=x+y$ and $x+y+z=1$, so we must show that \[(1-x)(1-y)(x+y)\left[\frac{x(1-x)}{(1+x)^2}+\frac{y(1-y)}{(1+y)^2}+\frac{2xy}{(1+x)(1+y)}\right]\le 4(3-2\sqrt 2)^2\] But from AM-GM we have $\frac{2xy}{(1+x)(1+y)}\le \frac{x^2}{(1+x)^2}+\frac{y^2}{(1+y)^2}$ and $(1-x)(1-y)\le \frac{(2-x-y)^2}{4}$. From the Jensen's inequality for the concave function $f:(0,1)\mapsto \Bbb{R},\,f(x)= \frac{x}{(1+x)^2}$ we have $\frac{x}{(1+x)^2}+\frac{y}{(1+y)^2}\le \frac{x+y}{(1+\frac{x+y}{2})^2}$. So, if we denote $u=\frac{x+y}{2}$, we must prove that \[\frac{(1-u)^2u^2}{(1+u)^2}\le (3-2\sqrt 2)^2\Longleftrightarrow \frac{u-u^2}{1+u}\le (\sqrt 2-1)^2\Longleftrightarrow (u-\sqrt 2+1)^2\ge 0\,.\] Equality hold when $x=y=\sqrt 2-1,\,z=3-2\sqrt 2\Longleftrightarrow \angle C=90^\circ, \angle A=\angle B=45^\circ$.

First Consider a other triangle $A'B'C'$ in a complex plane. Let $C'$ be the origin and $A'=a,B'=b,D'=z_1,E'=z_2$ Now easy to determine $z_1=\frac {b|a|}{|a|+|a-b|}$ and $z_2=\frac {a|b|}{|b|+|a-b|}$ Now letting $a=1$ we've $z_1-z_2=\frac {b}{1+|1-b|}-\frac {|b|}{|b|+|1-b|}$ Now suppose $|B|=1$ then we get |$z_1-z_2|=|\frac {b-1}{1+|1-b|}|$ So in this case now it's same as to show $\frac{2}{|b-1|}+|b-1|\geq 2\sqrt2$.... direct from $AM-GM$ Now so we need to look for case when triangle is not isosceles. Now it's clear that if $P,Q$ be points on $AB$ beyond $B$ and on $AC$ beyond $C$ then $PQ>BC$. Now after proving for the case isosceles consider the triangle now if we don't have $\angle {3C+2A}<180^0 $ or $\angle {3C+2B}<180^0 $ then not done directly , but since one of them is false so done.

I considered $D,E$ the feet of the bisectors of $B$ and $C$ by mistake. The solution (and problem) should nonetheless still be correct due to symmetry reasons. We have $AE=\dfrac{bc}{a+b},\ AD=\dfrac{bc}{b+c}$ so by the law of cosines in $\triangle{ADE}$ we get $$ DE^2=\dfrac{abc}{(a+b)^2(a+c)^2} \cdot (abc+(a^2-(b-c)^2) (a+b+c)) $$ We therefore have $DE^2\le \dfrac{abc}{(a+b)^2(a+c)^2}\cdot (abc+a^2(a+b+c))=\dfrac{a^2bc}{(a+b)(a+c)}$ or equivalently $$DE\le a\sqrt{\dfrac{bc}{(a+b)(a+c)}}$$ We know that $b+c\ge 2\sqrt{bc}$ and by Cauchy we know $(a+b)(a+c)\ge \left ( a+\sqrt{bc} \right )^2$ so it would be enough to prove that $$ \dfrac{a\sqrt{bc}}{a+\sqrt{bc}}\le (3-2\sqrt{2})(a+2\sqrt{bc})\Leftrightarrow a\sqrt{bc}\le (3-2\sqrt{2})(a^2+2bc+3a\sqrt{bc})$$By AM-GM, $a^2+2bc\ge 2\sqrt{2}\cdot a\sqrt{bc}$ so the inequality is proven. Equality holds for $b=c$ and $a=\sqrt{2bc}=\sqrt{2}b$ so for the right angled isosceles triangle.