(AI + DI)^2 + (CI + BI)^2 <= ( (AI^2 + DI^2)^1/2 + (BI^2 + CI^2)^1/2 )^2 by the Cauchy Schwarz Inequality (easily seen by expanding). However, AI*DI sin<AID = rAD, BI*CI sin(pi-<AID)=rBC where r is w's inradius. And by Cosine Law on AID, (AI^2 + DI)^1/2 = (AD^2 + 2AI*ID cos<AID)^1/2 = (AD^2 + 2rAD cos<AID/sin<AID)^1/2 <= AD + r cos<AID/sin<AID with equality iff cos <AID=0, i.e. <AID=pi/2. Similarly, we find that (BI^2 + CI^2)^1/2 <= BC + r cos(pi-<AID)/sin(pi-<AID) = BC - r cos <AID/sin<AID with equality iff <AID = pi/2, as before. Therefore, (AI + DI)^2 + (CI + DI)^2 <= ( (AI^2 + DI^2)^1/2 + (BI^2 + CI^2)^1/2 )^2 <= (AD + BC)^2 = (AB+CD)^2. Equality holds iff <AID = pi/2 and AI/BI=CI/DI (the Cauchy equality condition). i.e. Equality holds iff AID ~ BIC. Therefore, <DAB=<CBA, <ADC=<BCD from which it follows immediately that ABCD is an isosceles trapezoid.

There is a minor error above. The first inequality should be (AI + DI)^2 + (CI + BI)^2 ( (AI^2 + BI^2)^1/2 + (DI^2 + CI^2)^1/2 )^2. Then, instead of working with triangles AID and BIC, follow the exact same procedure with triangles AIB and CID. The same procedure and same conclusion will follow.

Alternately we can do a bash (though it is better than it looks)... let the length of the tangents to the incircle from $ A,B,C,D$ be $ a,b,c,d$ respectively. First we prove a lemma. Lemma: If $ r$ is the inradius of a tangential quadrilateral $ ABCD$ then \[ r^2 = \frac {abc + bcd + cda + dab}{a + b + c + d} = \frac {ab(c + d) + cd(a + b)}{a + b + c + d}.\] Proof: Let $ \alpha = \tan{A/2},\beta = \tan{B/2},\gamma = \tan{C/2},\delta = \tan{D/2}$. Letting $ s_1 = \sum\alpha$, $ s_2 = \sum\alpha\beta$, $ s_3 = \sum\alpha\beta\gamma$, and $ s_4 = \alpha\beta\gamma\delta$, we can prove by induction that \[ 0 = \tan\pi = \tan((A/2) + (B/2) + (C/2) + (D/2)) = \frac {s_1 - s_3}{1 - s_2 + s_4},\] so \[ s_1 = s_3,\] and the lemma follows.$ \blacksquare$ By Cauchy, \begin{align*} (AI + DI)^2 + (BI + CI)^2 & = AI^2 + BI^2 + CI^2 + DI^2 + 2(AI\cdot DI + BI\cdot CI) \\ & \le AI^2 + BI^2 + CI^2 + DI^2 + 2\sqrt {(AI^2 + BI^2)(DI^2 + CI^2)} \\ & = a^2 + b^2 + c^2 + d^2 + 4r^2 + 2\sqrt {(a^2 + b^2 + 2r^2)(c^2 + d^2 + 2r^2)}, \end{align*} with equality only when $ AI/DI = BI/CI$.* We claim that this is at most \[ (AB + CD)^2 = (a + b + c + d)^2 = a^2 + b^2 + c^2 + d^2 + 2(ab + ac + ad + bc + bd + cd),\] so we need to prove that \[ ab + ac + ad + bc + bd + cd\ge2r^2 + \sqrt {(a^2 + b^2 + 2r^2)(c^2 + d^2 + 2r^2)}.\] Using our lemma and the substitutions $ u = a + b$, $ v = ab$, $ w = c + d$, and $ x = cd$ for convenience (so $ r^2 = (ux + vw)/(u + w)$), this is equivalent to \[ (u + w)(v + x + uw) - 2(ux + vw) \\ \ge\sqrt {uw(u(u + w) - 2(v - x))(w(u + w) + 2(v - x))}\] or \begin{align*} (uw(u + w) + (u - w)(v - x))^2 & \ge uw(u(u + w) - 2(v - x))(w(u + w) + 2(v - x)) \\ u^2w^2(u + w)^2 + 2uw(u^2 - w^2)(v - x) + (u - w)^2(v - x)^2 & \ge u^2w^2(u + w)^2 + 2uw(u^2 - w^2)(v - x) - 4uw(v - x)^2 \\ (u + w)^2(v - x)^2 & \ge0, \end{align*} as desired. Equality occurs when $ ab = cd$, so \[ r^2 = \frac {ab(c + d) + cd(a + b)}{a + b + c + d} = \frac {ab(a + b + c + d)}{a + b + c + d} = ab = cd.\] But this can only be true if (note similar triangles with $ r/a = b/r$ and so on) \[ \angle AIB = \angle CID = 90^\circ.\] Now from *, we have $ AI/DI = BI/CI$, so $ \triangle AIB$ and $ \triangle CID$ are similar right triangles with the same length $ r$ of altitude to the hypotenuse, so they are congruent, and $ a = d,b = c$ so $ a + b = c + d$ and we conclude that $ ABCD$ is an isosceles trapezoid.

Here's a more trigonometric solution... Without loss of generality, let $\omega$ have radius 1. Let $\omega$ be tangent to $AB$, $BC$, $CD$, and $DA$ at $W$, $X$, $Y$, and $Z$, respectively. It is clear that $AI$ bisects $\angle ZIW$, $BI$ bisects $\angle WIX$, etc. Let $m \angle AIW = 2a$, $m \angle WIX = 2b$, $m \angle XIY = 2c$, $m \angle YIZ = 2d$. By some simple trigonometry, $AI = \sec a$, $BI = \sec b$, $CI = \sec c$, $DI = \sec d$, and $AB + CD = \tan a + \tan b + \tan c + \tan d$. Furthermore, $a,b,c,d < \frac{\pi}{2}$ and $a+b+c+d = \pi$. We wish to show that $a=b$ and $c=d$. The given equation can be rewritten as $(\sec a + \sec d)^2 + (\sec b + \sec c)^2 = (\tan a + \tan b + \tan c + \tan d)^2$. Expanding, $\sec^2 a + \sec^2 b + \sec^2 c + \sec^2 d + 2 \sec a \sec d + 2 \sec b \sec c = (\sec^2 a - 1) + (\sec^2 b - 1) + (\sec^2 c - 1) + (\sec^2 d - 1) + 2 (\tan a \tan b + \tan a \tan c + \tan a \tan d + \tan b \tan c + \tan b \tan d + \tan c \tan d)$. Rearrange this to get $\sec a \sec d + \sec b \sec c + 2 = \tan a \tan d + \tan b \tan c + (\tan a + \tan d)(\tan b + \tan c)$, or $(\sec a \sec d - \tan a \tan d + 1) + (\sec b \sec c - \tan b \tan c + 1) = (\tan a + \tan d)(\tan b + \tan c)$. This rearranges to ${\frac{1 - \sin a \sin d + \cos a \cos d}{\cos a \cos d} + \frac{1 - \sin b \sin c + \cos b \cos c}{\cos b \cos c} = (\frac{\sin a \cos d + \sin d \cos a}{\cos a \cos d}}) ( \frac{\sin b \cos c + \sin c \cos b}{\cos b \cos c} ),$ which rearranges again to $\frac{1 + \cos(a+d)}{\cos a \cos d} + \frac{1 + \cos(b+c)}{\cos b \cos c} = \frac{\sin(a+d) \sin(b+c)}{\cos a \cos b \cos c \cos d}.$ Clearing denominators, we get that $(\cos b \cos c)(1 + \cos(a+d)) + (\cos a \cos d)(1 + \cos(b+c)) = \sin(a+d) \sin(b+c).$ Note that $f(x) = \log \cos x$ over $0 < x < \frac{\pi}{2}$ has $f''(x) = - \sec^2 x < 0$, so $f$ is concave. By Jensen's inequality, it follows that $\log \cos \frac{x+y}{2} \geq \frac{\log \cos x + \log \cos y}{2}$, that is, $\cos x \cos y \leq \cos^2 \frac{x+y}{2}$ for any $x,y \in (0, \frac{\pi}{2})$, with equality holding iff $x=y$. Since $1 + \cos(a+d), 1 + \cos(b+c) \geq 0$, it follows that $\sin(a+d) \sin(b+c) = (\cos b \cos c)(1 + \cos(a+d)) + (\cos a \cos d)(1 + \cos(b+c)) \leq \cos^2 \frac{b+c}{2} (1 + \cos(a+d)) + \cos^2 \frac{a+d}{2} (1 + \cos(b+c))$. Let $\alpha = a+d$ and let $\beta =b+c$; then $\alpha+\beta = \pi$. We have that $\sin \alpha \sin \beta \leq \cos^2 \frac{\beta}{2} (1 + \cos \alpha) + \cos^2 \frac{\alpha}{2} (1 + \cos \beta)$ with equality holding iff $a=d$ and $b=c$. On the other hand, $\begin{align*}\cos^2 \frac{\beta}{2} (1 + \cos \alpha) + \cos^2 \frac{\alpha}{2} (1 + \cos \beta) &= \frac{1 + \cos \beta}{2} (1 + \cos \alpha) + \frac{1 + \cos \alpha}{2} (1 + \cos \beta) \\ &= (1 + \cos \beta)(1 + \cos \alpha) \\ &= (1 + \cos \beta)(1 + \cos(\pi - \beta)) \\ &= 1 - \cos^2 \beta \\ &= \sin^2 \beta \\ &= \sin \alpha \sin \beta,\end{align*},$ so equality always holds. Hence, we must have that $a=d$ and $b=c$, which completes our proof.

Here's a completely algebraic solution. WLOG $\omega$ has radius $1$, and let $a$, $b$, $c$, $d$ be the lengths of the tangents from $A$, $B$, $C$, $D$ to $\omega$. It follows that $a+b+c+d = abc+bcd+cda+dab$ ($\star$) as many people noted. Then, the content of the problem is to show that \[ (\sqrt{a^2+1}+\sqrt{d^2+1})^2 + (\sqrt{b^2+1}+\sqrt{c^2+1})^2 \le (a+b+c+d)^2 \]subject to ($\star$), with equality only when $a=d=\tfrac1b=\tfrac1c$. Let $S = ab+bc+cd+da+ac+bd$. Then the inequality is \[ \sqrt{(a^2+1)(d^2+1)} + \sqrt{(b^2+1)(c^2+1)} \le S - 2. \]Now, by USAMO 2014 Problem 1 and the condition ($\star$), we have that $(a^2+1)(b^2+1)(c^2+1)(d^2+1) = (S - abcd - 1)^2$. So squaring both sides, the inequality becomes \[ (ad)^2+(bc)^2 + a^2+b^2+c^2+d^2 \le S^2 - 6S + 2abcd + 4. \]To simplify this, we use the identities \begin{align*} S^2 &= 6abcd + \sum_{\text{sym}} a^2bc + \frac14\sum_{\text{sym}} a^2b^2 \\ (a+b+c+d)^2 &= (abc+bcd+cda+dab)(a+b+c+d) \\ &= 4abcd + \frac12\sum_{\text{sym}} a^2bc \end{align*}So $S^2+2abcd = \tfrac14\textstyle\sum_{\text{sym}} a^2b^2 + 2(a^2+b^2+c^2+d^2) + 4S$ and the identity we want to prove reduces to \[ 2S \le (ab)^2+(ac)^2+(bd)^2+(cd)^2 + 4 + a^2 + b^2 + c^2 + d^2. \]This is immediate by repeated applications of AM-GM, and the equality case is when $ab=ac=bd=cd=1$, $a=d$, $b=c$.

Construct a point $J$ such that $JBC\sim IAD$ and $JBCI$ is cyclic. Let the ratio of similarity $\frac{AD}{BC}$ be $r$, and let $JB=a,JC=b, IB=c,IC=d$. Note that $AI=ra$ and $BI=rb$. Additionally, \[ r = \frac{AD}{BC} = \frac{[IAD]}{[IBC]} = r^2\frac{[JBC]}{[IBC]}= r^2 \frac{JB\cdot JC \sin \theta}{IB\cdot IC \sin \theta} = r^2 \frac{ab}{cd} \]so $r = \frac{cd}{ab}$. Finally, it's well known that $AB+CD=AD+BC$. So the condition reduces to \[ (ra+rb)^2 + (c+d)^2 = (AD+BC)^2 = (1+r)^2 BC^2 \]or equivalently \[ c^2d^2(a+b)^2+a^2b^2(c+d)^2=(ab+cd)^2 BC^2 \]Now a few more computations yield $BC^2 = \frac{(ac+bd)(ad+bc)}{ab+cd}$ so the condition is \begin{align*} c^2d^2(a+b)^2+a^2b^2(c+d)^2&=(ab+cd)(ac+bd)(ad+bc)\\ \Leftrightarrow abcd(a^2+b^2-2ab+c^2+d^2-2ac)&=0\\ \Leftrightarrow a=b, c=d. \end{align*}

math154 wrote: Lemma: If $ r$ is the inradius of a tangential quadrilateral $ ABCD$ then \[ r^2 = \frac {abc + bcd + cda + dab}{a + b + c + d} = \frac {ab(c + d) + cd(a + b)}{a + b + c + d}.\]Proof: Let $ \alpha = \tan{A/2},\beta = \tan{B/2},\gamma = \tan{C/2},\delta = \tan{D/2}$. Letting $ s_1 = \sum\alpha$, $ s_2 = \sum\alpha\beta$, $ s_3 = \sum\alpha\beta\gamma$, and $ s_4 = \alpha\beta\gamma\delta$, we can prove by induction that \[ 0 = \tan\pi = \tan((A/2) + (B/2) + (C/2) + (D/2)) = \frac {s_1 - s_3}{1 - s_2 + s_4},\]so \[ s_1 = s_3,\]and the lemma follows.$ \blacksquare$ Why does this prove the lemma?

pi37 wrote: Construct a point $J$ such that $JBC\sim IAD$ and $JBCI$ is cyclic \end{align*} How can we be sure such a point even exists?

Sorry for coming to a old problem but is it possible to solve with complex numbers in a elegant way? As the solutions are mainly a bunch of calculations I asking myself if using complex is a good idea.

Let $a,b,c,d$ denote the length of the tangents from $A,B,C,D$, respectively, and let $\angle AID=\theta$ As others have noted above, we have $$r^2=\frac{abc+bcd+cda +dab}{a+b+c+d}$$Note that we easily get $\angle BIC=180^{\circ}-\theta$ from angle chasing. The problem falls to the following: Claim: $AD\cdot BC=4r^2$ Proof: We apply Law of Cosines on $\triangle AID$ and $BIC$. Plugging this into the condition we get $$(AI+DI)^2+(BI+CI)^2=AD^2+BC^2 +2AI\cdot DI(\cos\theta +1)+2BI\cdot CI(1-\cos\theta)=(AB+CD)^2=(AD+BC)^2 $$where the last equality comes from Pitot’s theorem. Rearranging, applying reverse power reduction on cosine, and using trig in right triangles gives $$AI\cdot DI(\cos\theta+1)+BI\cdot CI(1-\cos\theta)=AD\cdot BC$$$$\implies 2AI*DI\cdot \cos^2\frac{\theta}{2}+ 2BI*CI\cdot \sin^2\frac{\theta}{2}=AD\cdot BC$$$$\implies 2r^2+2r^2=4r^2=AD\cdot BC$$hence the claim is proven. We can use the radius formula to get $$4\cdot \frac{abc+bcd+cda +dab}{a+b+c+d}=(a+d)(b+c)$$Rearranging and completing squares gives $$(b+c)(a-d)^2+(a+d)(b-c)^2=0\implies a=d, b=c$$It is easy to see that this implies $ABCD$ is an isosceles trapezoid, as desired.

Let $\omega$ have radius $1$. Let $a=\sqrt{AI^2-1},b=\sqrt{BI^2-1},c=\sqrt{CI^2-1},d=\sqrt{DI^2-1}$. Then we have \[a^2+2\sqrt{a^2+1}\sqrt{d^2+1}+d^2+b^2+2\sqrt{b^2+1}\sqrt{c^2+1}+4=(a+b+c+d)^2.\]By cancelling, we have \[\sqrt{a^2+1}\sqrt{d^2+1}+\sqrt{b^2+1}\sqrt{c^2+1}=ab+bc+cd+da+ac+bd-2.\]At this point, note that by the tangent addition formula we have \[\frac{a+b}{1-ab}=-\frac{c+d}{1-cd}\implies a+b-acd-bcd=-c-d+cab+dab\implies a+b+c+d=abc+bcd+cda+dab.\]Let $ab+bc+cd+da+ac+bd=S$. Then, the key identity from USAMO 2014/1 is that \[(a^2+1)(b^2+1)(c^2+1)(d^2+1)=(S-abcd-1)^2.\]Now, square to yield \[a^2d^2+b^2c^2+a^2+b^2+c^2+d^2+2+2[S-abcd-1]=S^2-4S+4\implies\]\[a^2d^2+b^2c^2+a^2+b^2+c^2+d^2=S^2-6S+4+2abcd.\]Let $T=a+b+c+d$ and rearrange this as \[a^2d^2+b^2c^2-2abcd=(S-2)^2-T^2 = \]\[\frac 12\sum_{\mbox{sym}}a^2b^2 + \sum_{\mbox{sym}}a^2bc+6abcd-2\sum_{\mbox{sym}}ab+4-\sum_{\mbox{cyc}}a^2-(a+b+c+d)(abc+bcd+cda+dab)\iff\]\[(a^2+d^2)(b^2+c^2)+\frac 12\sum_{\mbox{sym}}a^2bc+4abcd-2\sum_{\mbox{sym}}ab+4=0.\]Rewrite this last equality as \[0=(ac-1)^2+(ab-1)^2+(cd-1)^2+(bd-1)^2+\frac 12 \sum_{\mbox{sym}}a^2bc + 4abcd-4ad-4bc-2(a+d)(b+c).\]So it is sufficient to check \[\frac 12 \sum_{\mbox{sym}}a^2bc + 4abcd\ge 4ad+4bc+2(a+d)(b+c)\]and find the equality case(s). Multiplying one side by $a+b+c+d$ and the other by $abc+bcd+cda+dab$ yields that we want to check \[7\sum_{\mbox{cyc}}a^2bcd+\frac 12\sum_{\mbox{sym}}a^3bc+\sum_{\mbox{sym}}a^2b^2c\ge 2[a^2bcd+abcd^2+ab^2cd+abc^2d+cd^2a^2+d^2a^2b+ab^2c^2+db^2c^2]+\]\[\sum_{\mbox{sym}}a^2b^2c+6\sum_{\mbox{cyc}}a^2bcd.\]Cancelling, we need to check \[\frac 12\sum_{\mbox{sym}}a^3bc\ge (a+b+c+d)abcd+2[cd^2a^2+d^2a^2b+ab^2c^2+db^2c^2].\]By the inequalities $abd(a-d)^2\ge 0,acd(a-d)^2\ge 0,abc(b-c)^2\ge 0,bcd(b-c)^2\ge 0$, it suffices to show \[a^3bc+b^3ad+c^3ad+d^3bc\ge (a+b+c+d)abcd.\]Finally, use cyclic versions of \[\frac{2a^3bc}{3}+\frac{d^3bc}{3}\ge a^2bcd\]to finish. Then, we extract that the equality case satisfies $a=d,b=c,ac=1$. Hence the tuple is $(a,b,c,d)=(a,\tfrac 1a,\tfrac 1a,a)$. This implies the desired.

Here is a somewhat different algebraic solution: WLOG the inradius is $1$. We are given $$(\sqrt{w^2 + 1} + \sqrt{z^2 + 1})^2 + (\sqrt{x^2 + 1} + \sqrt{y^2 + 1})^2 = (w+x+y+z)^2,$$where $w,x,y,z$ are positive reals and we have to prove $w = z, x = y$ and $wx = 1$. Expanding, we get $$4 + 2\sqrt{w^2 + 1}\sqrt{z^2 + 1} + 2\sqrt{x^2 + 1}\sqrt{y^2 + 1} = (w+x+y+z)^2 - (w^2 + x^2 + y^2 + z^2).$$Using the fact that $\sqrt{w^2 + 1}\sqrt{z^2 + 1} = \sqrt{(wz - 1)^2 + (w+z)^2}$ we can rewrite the new equation as: $$4 + 2\sqrt{(wz - 1)^2 + (w+z)^2} + 2\sqrt{(xy - 1)^2 + (x+y)^2} = (w+x+y+z)^2 - (w^2 + x^2 + y^2 + z^2).$$Also, the condition is: $$\frac{wz - 1}{w+z} = \frac{1 - xy}{x + y} = K$$where WLOG $K \ge 0$ (otherwise, we can switch the pair (w,z) and (x,y)). Now, let $u_1 = w + z, u_2 = wz, v_1 = x + y, v_2 = xy.$ By the condition, we get that $u_2 = 1 + Ku_1$ and $v_2 = 1 - Kv_1$. Substituting into the equation, we get: $$(w+x+y+z)^2 = (u_1 + v_1)^2,$$$$2\sqrt{(wz - 1)^2 + (w+z)^2} = 2u_1\sqrt{K^2 + 1},$$$$2\sqrt{(xy - 1)^2 + (x+y)^2} = 2v_1\sqrt{K^2 + 1},$$and $$w^2 + x^2 + y^2 + z^2 = u_1^2 - 2u_2 + v_1^2 - 2v_2 = u_1^2 + v_1^2 - 4 - 2K(u_1 - v_1).$$The equation can now be rewritten as: $$4 + 2u_1\sqrt{K^2 + 1} + 2v_1\sqrt{K^2 + 1} = 2u_1v_1 + 4 + 2K(u_1 - v_1),$$which can be simplified further: $$u_1v_1 + K(u_1 - v_1) = (u_1 + v_1)\sqrt{K^2 + 1}.$$We use SFFT here to write the equation as: $$(u_1 - (\sqrt{K^2 + 1} + K))(v_1 - (\sqrt{K^2 + 1} - K)) = 1.$$ Now observe that $u_1^2 = (w+z)^2 \ge 4wz = 4u_2 = 4 + 4Ku_1$. Similarly, $v_1^2 \ge 4 - 4Kv_1$. So, $$(u_1 - 2K)^2 \ge 4K^2 + 4 \implies |u_1 - 2K| \ge 2\sqrt{K^2 + 1}$$and similarly $$|v_1 + 2K| \ge 2\sqrt{K^2 + 1}.$$Since $u_1$ and $v_1$ are both positive, the absolute value signs cannot change any values, so $u_1 \ge 2(\sqrt{K^2 + 1} + K)$ and $v_1 \ge 2(\sqrt{K^2 + 1} - K)$. Using our SFFT equation, we check that equality actually holds everywhere, so $u_1v_1 = 4, u_1^2 = 4u_2,$ and $v_1^2 = 4v_2$. Thus $w = z, x = y, $ and $wx = 1$ as desired.

WLOG inradius $r = 1$.Let $AC \cap BD = X$ (WLOG $X$ lies on $\overrightarrow{AB}$), $AB,BC,CD,DA$ tangent to $\omega$ at $P,Q,R,S$ and the length of tangent line from $A,B,C,D,X$ to $\omega$ are $a,b,c,d,l$ respectively. Let $p,q$ be the length of $AD,BC$ respectively ($p = a+d,q=b+c$). By well-known $[\triangle] = rs$, $r^2 = 1 = \frac{lad}{l+a+d} = \frac{lbc}{l-b-c}$. So, $ad = \frac{l+p}{l} = 1+\frac{p}{l}$. Similarly, $bc = 1-\frac{q}{l}$. By the problem equation, \begin{align*} a^2+b^2+c^2+d^2 +4 + 2(\sqrt{(a^2+1)(d^2+1)} + \sqrt{(b^2+1)(c^2+1)}) &= (p+q)^2 \\ \iff p^2-2(1+\frac{p}{l}) + q^2 - 2(1-\frac{q}{l}) +4 + 2(\sqrt{(1+\frac{p}{l})^2 + p^2 - 2(1+\frac{p}{l}) +1} + \sqrt{(1-\frac{q}{l})^2 + q^2 - 2(1-\frac{q}{l}) +1}) &=p^2+2pq+q^2\\ \iff p = q \left( \frac{1+\sqrt{1+l^2}}{1-\sqrt{1+l^2}+ql}\right)

\end{align*}Note that $p,q$ obtain the minimum value when $X,I,Q$ and $X,I,S$ are collinear. So $p \geq \frac{2(1+\sqrt{l^2+1})}{l}$ and $q \geq \frac{2(-1+\sqrt{l^2+1})}{l}$Proof

. Consider function $f(x) = \frac{x}{1-\sqrt{1+l^2}+xl}$. For $q_1>q_2 \geq \frac{2(-1+\sqrt{l^2-1})}{l}$, we get $f(q_1)-f(q_2) = \frac{(q_1-q_2)(1-\sqrt{1+l^2})}{(1-\sqrt{1+l^2}+q_1l)(1-\sqrt{1+l^2}+q_2l)}<0$. Therefore, $f$ is deceasing function. Hence, $$ \frac{2(1+\sqrt{l^2+1})}{l} \leq p = f(q) \leq f\left(\frac{2(-1+\sqrt{l^2+1})}{l} \right) = \frac{2(1+\sqrt{l^2+1})}{l}.$$The equality holds when $X,I,Q$ and $X,I,P$ are collinear. Therefore, $ABCD$ is an isosceles trapezoid.

Suppose the tangents to the incircle from $A$, $B$, $C$, $D$ have lengths $a$, $b$, $c$, $d$, respectively, and let the inradius equal $r$. Recall \[r = \sqrt{\frac{abc+bcd+cda+dab}{a+b+c+d}}.\] We begin by manipulating the LHS of our inequality using Cauchy: \begin{align*} \text{LHS} &= \left(\sum_{\text{cyc}} AI^2\right) + 2(AI \cdot DI + BI \cdot CI) \\ &\leq \left(\sum_{\text{cyc}} AI^2\right) + 2\sqrt{(AI^2+BI^2)(DI^2+CI^2)} \\ &= \left(\sum_{\text{cyc}} a^2\right) + 4r^2 + 2\sqrt{(a^2+b^2+2r^2)(c^2+d^2+2r^2)}. \end{align*} We claim the RHS is greater than or equal to the previous expression. Indeed, expanding gives us \[0 \leq (ab-cd)^2(a+b+c+d)^2,\] so we must have the equality case $ab=cd$ as well as $\frac{AI}{DI} = \frac{BI}{CI}$ from Cauchy. Substituting the first equality back into our formula for $r$, we get $r^2 = ab = cd$, so \[\angle AIB = \angle DIC = 90 \implies \triangle AIB \sim \triangle DIC\] in combination with our ratio condition. Moreover, the corresponding altitudes from $I$ are the same length $r$, giving congruence and thus the length equality $AB = CD$. To finish, we see this similarity allows to easily find that the altitudes from $I$ to $AD$, $BC$ form a 180 degree angle. Hence $AD \parallel BC$, forming our desired isosceles trapezoid. $\blacksquare$