I have never known, and i ahve asked thousands of times but i never remember, what is a circumscribed cuadrilateral? I know it is A or B but i dont know wich..... Pascual

well it's the exact opposite of cyclic quadrilaterals: it's a quadrialetaral with a circle inside, which touches all the sides. Obviously the sum of the opposite sides is the same (because they are composed of 4 segments each being a tangent to the circle from one of the 4 vertices).

hey, can anyone post an elegant solution for this one, I foun a solution but it is horrible! and also full of details i have to improve, so ANY other solution would be welcome......

um ok. this solution may in fact be the "horrible solution" that you found but anyway, here goes: let AB=a, BC=b, CD=c, AD=d. note it is sufficient to prove one side of the inequality, since the other follows directly by symmetry. So we want to prove: 1/3 (a^3-d^3) leq (b^3-c^3). note we can WLOG a>d, b>c, since a+c=b+d, a-d=b-c and thus a>d, b<c is contradiction. a-d=b-c, so that inequality is equivalant to (a^2+d^2+ad) <= 3(b^2+c^2+bc). since b^2+c^2+bc=a^2-2ad+d^2+3bc, that ienquality is equivalant to : 7ad <= 2a^2+2d^2+9bc. since 2a^2+2d^2 >= 4ad, it would now be sufficient to prove 3ad <= 9bc, i.e. 3bc >= ad (which is in fact true). My way of proving this is a ruthless bash: let the angle bisectors meet at the incentre I of ABCD. Let <IAB=<IAD=p, <ABI=<CBI=q, <BCI=<DCI=r, <CDI=<ADI=s, inradius = n. Note that 30 < p,q,r,s < 60, so 1/sqrt 3 < tan p, etc < sqrt 3. since a = n (tan p + tan q), b = n(tan q + tan r), etc; proving 3bc >= ad is equivalant to proving that (tan r + tan s)(tan q + tan r)/(tan p + tan q)(tan p + tan s) <= 3. now, we know that r+s+p+q=180, so using the horrible tan sum formula, we can deduce that the sum of the numbers tan p, tan q, tan r and tan s is equal to the sum of their product in threes. after using *lots* of algebra, we can simplify that expression to something much nicer : (tan r)^2+1 / (tan p)^2 + 1. this obviously has a maximum of 3, attained when r = 60, p = 30. so we are done. the condition for equality occurs when: firstly, 2a^2+2d^2 = 4ad, and 3bc = ad, i.e. a = d, r = 60 and p = 30 (which is a very wierd condition for equality).

I have this solution: Note first that it suffices to show only one of the inequalities, since we can obtain one from another interchanging vertices $A$ and $C$. Since the quadrilateral is circumscribed, $AB - AD = BC - CD$. Thus the inequality is equivalent to \[\frac13(AB^2 + AB\cdot AD + AD^2) \leq BC^2 + BC\cdot CD + CD^2\] The condition concerning angles on the quadrilateral is equivalent to its internal angles being between $60^\circ$ and $120^\circ$. Hence, by the law of cosines, \[ BC^2 + BC\cdot CD + CD^2 = BC^2 + CD^2 - 2\cdot BC\cdot CD\cdot \cos 120^\circ \] \[ \geq BC^2 + CD^2 - 2\cdot BC\cdot CD\cdot \cos C = BD^2\] Thus we just have to show that $BD^2 \geq \frac13(AB^2 + AB\cdot AD + AD^2)$. But $BD^2 = AB^2 + AD^2 - 2\cdot AB\cdot AD\cos A$, hence, since \[ AB^2 + AD^2 - 2\cdot AB\cdot AD\cos A\geq \frac13(AB^2 + AB\cdot AD + AD^2) \] \[ \iff 2AB^2 + 2AD^2 \geq AB\cdot AD(1 + 6\cos A)\] and $\cos A \leq \frac12$, \[ AB(1 + 6\cos A) \leq 4AB \leq 2AB^2 + 2AD^2,\] because $AB^2 + AD^2 \leq 2AB\cdot AD$ and the result follows.

Couldn't you use rearrangement...?

cyshine wrote: Since the quadrilateral is circumscribed, $ AB - AD = BC - CD$. Why? I think I misunderstand "circumscribed quadrilateral", can someone make a dwawing, in particular showing what "exterior angles" are here?

nice one!

This is pretty much the same as cyshine's but with more explanation. We set $ a = BC$, $ b = CD$, $ c = DA$, and $ d = AB$. By a well-known property of tangential quadrilaterals, the sum of the two pairs of opposite sides are equal; hence $ a + c = b + d \Rightarrow a - b = d - c \Rightarrow |a - b| = |d - c|$ Now we factor the desired expression into $ \frac {|d - c|(c^2 + d^2 + cd)}{3} \le|a - b|(a^2 + b^2 + ab)\le 3|d - c|(c^2 + d^2 + cd)$. Temporarily discarding the case where $ a = b$ and $ c = d$, we can divide through by the $ |a - b| = |d - c|$ to get the simplified expression $ (c^2 + d^2 + cd)/3\le a^2 + b^2 + ab\le 3(c^2 + d^2 + cd)$. Now, draw diagonal $ BD$. By the law of cosines, $ c^2 + d^2 + 2cd\cos A = BD$. Since each of the interior and exterior angles of the quadrilateral is at least 60 degrees, we have that $ A\in [60^{\circ},120^{\circ}]$. Cosine is monotonously decreasing on this interval, so by setting $ A$ at the extreme values, we see that $ c^2 + d^2 - cd\le BD \le c^2 + d^2 + cd$. Applying the law of cosines analogously to $ a$ and $ b$, we see that $ a^2 + b^2 - ab\le BD \le a^2 + b^2 + ab$; we hence have $ c^2 + d^2 - cd\le BD \le a^2 + b^2 + ab$ and $ a^2 + b^2 - ab\le BD \le c^2 + d^2 + cd$. We wrap up first by considering the second inequality. Because $ c^2 + d^2 - cd\le BD \le a^2 + b^2 + ab$, $ \text{RHS}\ge 3(a^2 + b^2 - ab)$. This latter expression is of course greater than or equal to $ a^2 + b^2 + ab$ because the inequality can be rearranged to $ 2(a - b)^2\ge 0$, which is always true. Multiply the first inequality by $ 3$ and we see that it is simply the second inequality with the variables swapped; hence by symmetry it is true as well. Equality occurs when $ a = b$ and $ c = d$, or when $ ABCD$ is a kite.

Since quadrilateral $ABCD$ is circumscribed about a circle we have $AB+CD=BC+AD \Rightarrow |AB-AD|=|BC-CD|$.Let us consider the case when $AB \neq AD$ and $BC \neq CD$.Then the inequality is equivalent to $\frac{1}{3}(AB^2+AD^2+AB*AD) \le BC^2+CD^2+BC*CD \le 3(AB^2+AD^2+AB*AD)$ Before giving the proof I first provide a prerequisite lemma: Lemma:$\frac{BC*CD}{AB*AD}=\frac{1-cosA}{1-cosC}$ Proof: $BD^2=AB^2+AD^2-2AB*ADcosA=BC^2+CD^2-2BC*CDcosC \Rightarrow AB*AD(1-cosA)=BC*CD(1-cosC) \Rightarrow \frac{BC*CD}{AB*AD}=\frac{1-cosA}{1-cosC}$ We first show the left part of the inequality,which is equivalent to $((AB-AD)^2+3AB*AD) \le 3((BC-CD)^2+3BC*CD)$ $\Leftrightarrow 2(AB-AD)^2+9BC*CD-3AB*AD \ge 0$ Now $9BC*CD-3AB*AD \ge 0$ $\Leftrightarrow 3BC*CD-AB*AD \ge 0$ $\Leftrightarrow 3\frac{BC*CD}{AB*AD}-1 \ge 0$ $\Leftrightarrow 3\frac{1-cosA}{1-cosC}-1 \ge 0$ $\Leftrightarrow 3cosA-cosC \le 2$ Now $60^{\circ} \le \angle{A},\angle{C} \le 120^{\circ} \Rightarrow cosA \le \frac{1}{2},cosC \ge \frac{-1}{2}$ from which the above inequality follows.Thus with this the left part of the inequality follows. Similar simplications of the rightpart suffices to prove that $3AB*AD \ge BC*CD$ for which the proof is similar to the left part. In the case when $AB=AD$ we get $BC=DC$ for which the equality holds(note that otherwise there is strict inequality)

Silly problem... Clearly it suffices to show the left inequality. Since $AB+CD = BC+AD \implies |AB-AD| = |BC-CD|$, it suffices to prove \[ \frac13(AB^2 + AB \cdot AD + AD^2) \le BC^2 + BC \cdot CD + CD^2. \]This follows by noting that \begin{align*} BC^2 + BC \cdot CD + CD^2 &\ge BC^2 + CD^2 - 2(BC)(CD)\cos(\angle BCD) \\ &= BD^2 \\ &= AB^2 + AD^2 - 2(AB)(AD)\cos(\angle BAD) \\ &\ge AB^2 + AD^2 - AB \cdot AD \\ &\ge \tfrac13(AB^2 + AD^2 + AB \cdot AD) \end{align*}the last line following by AM-GM.

WLOG the incircle of $ABCD$ has radius $1$. Let $\alpha = A/2$, etc. It suffices to prove the LHS, and then the RHS will follow by symmetry of applying the LHS result with $ABCD$ changed to $BCDA$. After some manipulations we find that the inequality is equivalent to $(\cot \alpha + \cot \beta)^2 + (\cot \alpha + \cot \delta)^2 + (\cot \alpha + \cot \beta)(\cot \alpha + \cot \delta) \le 3((\cot \gamma + \cot \beta)^2 + (\cot \gamma + \cot \delta)^2 + (\cot \gamma + \cot \beta)(\cot \gamma + \cot \delta))$. Translate everything back to $AB, BC, CD,DA$ it suffices to prove that $$AB^2 + AD^2 + AB\cdot AD\le 3(BC^2 + CD^2 + BC\cdot CD).$$ Indeed, observe that $3(BC^2 + BD^2+BC \cdot CD) \ge 3BD^2 \ge 3(AB^2 + AD^2 - AB\cdot AD) \ge \frac{3(AB^2 + AD^2)}{2} \ge AB^2 + AD^2 + AB\cdot AD$. Edit: whoops shouldn't have substituted

WLOG, we only need to prove one of the inequalities. Note that $AB-AD=BC-CD$ by Pitot's. Note that $AB^2+AB\cdot AD+AD^2\ge AD^2$ because $\angle BAD\le 120^\circ$ and $BC^2+CD^2-BC\cdot CD\le AD^2$ because $\angle BCD\ge 60^\circ.$ Thus, we have $AB^2+AB\cdot AD+AD^2\ge BC^2-BC\cdot CD+CD^2.$ It remains to prove $3(BC^2-BC\cdot CD+CD^2)\ge BC^2+BC\cdot CD+CD^2$ which is true.

Nice geo ineq! Clearly, by symmetry, we only need to show one side. Let's show the left side. By Pitot, $|BC-CD|=|AB-AD|$, so we only need to show that $$3(BC^2+BC\cdot CD+CD^2)\geq BA^2+BA\cdot AD+AD^2.$$ Note that $$3(BC^2+BC\cdot CD+CD^2)= 3(BC^2+CD^2-2\cdot BC\cdot CD\cos{120})\geq 3BD^2.$$We also have $$3BD^2\geq 3(BA^2+AD^2-2BA\cdot AD\cos(60))=3BA^2+3AD^2-3BA\cdot AD$$We then wish to show that $$3BA^2+3AD^2-3BA\cdot AD\geq BA^2+BA\cdot AD+AD^2,$$but this just rearranges to $$BA^2+AD^2\geq 2BA\cdot AD,$$which is clearly true, so we are done. Remarks: When I first tried this problem a few months ago, I failed. What I did back then was trying to fix the inradius as 1, letting the incircle tangency points to $AB,BC,CD,DA$ be $W,X,Y,Z$ respectively, and let $AW=AZ=a,BW=BX=b$, and so on. The angle condition becomes $1/\sqrt{3}\leq a,b,c,d\leq \sqrt{3}$, and then I tried to expand everything and prove the resulting inequality. However, the issue with this approach is that $a,b,c,d$ are not actually independent, and the resulting inequality is not true for all $1/\sqrt{3}\leq a,b,c,d\leq \sqrt{3}$, which forces you to use the very cumbersome condition $$\arctan{a}+\arctan{b}+\arctan{c}+\arctan{d}=180,$$which is hard to incorporate into the inequality.

By symmetry, we only need to show the first inequality. We may also assume WLOG that $AB>AD$ and $BC>CD$. Then simply notice the inequality $$\frac 13(AB^2+AB \cdot AD + AD^2) \leq AB^2 - AB \cdot AD + AD^2 \leq BD^2 \leq BC^2+BC \cdot CD+CD^2$$by Law of Cosines and AM-GM, as needed.

By symmetry it suffices to show the left inequality. Note that $$BD^2 = BC^2 + CD^2 + 2BC\cdot CD \cos \angle BCD \leq BC^2 + BC \cdot CD + CD^2.$$By AM-GM, we also have $$\frac{1}{3}(AB^2 + AD^2 + AB \cdot AD) \leq AB^2 + AD^2 - AB \cdot AD \leq AB^2 + AD^2 + 2 \cos \angle DAB = BD^2,$$so we get $$\frac{1}{3}(AB^2 + AD^2 + AB \cdot AD) \leq BD^2 \leq BC^2 + CD^2 + BC \cdot CD.$$Multiplying both sides by $|AB-AD| = |BC - CD|$ we get $$\frac{1}{3}|AB^3-AD^3| \leq |BC^3 - CD^3|,$$as desired.