# For $a$, $b$, $c$, $d$ the sides of a quadrilateral, show $ab^2(b-c)+bc^2(c-d)+cd^2(d-a)+da^2(a-b)\ge 0$. (A generalization of IMO 1983 problem 6)

Let $$a$$, $$b$$, $$c$$, and $$d$$ be the lengths of the sides of a quadrilateral. Show that $$ab^2(b-c)+bc^2(c-d)+cd^2(d-a)+da^2(a-b)\ge 0 \tag{\star}$$

Background: The well known 1983 IMO Problem 6 is the following:

IMO 1983 #6. Let $$a$$, $$b$$ and $$c$$ be the lengths of the sides of a triangle. Prove that $$a^{2}b(a - b) + b^{2}c(b - c) +c^{2}a(c - a)\ge 0.$$

See: here. A lot of people have discussed this problem. So this problem (IMO 1983) has a lot of nice methods.

Now I found for quadrilaterals a similar inequality. Are there some nice methods for inequality $$(\star)$$?

• Are you certain that the inequality for quadrilateral is correct? – zyx Aug 13 '13 at 0:48
• I'm sure, I have use bottem test is ture – math110 Aug 13 '13 at 0:50
• What is bottem test? – zyx Aug 13 '13 at 0:53
• old.irgoc.org/Soft/ShowSoft.asp?SoftID=15， This software is the best software validation inequality, developed by several Chinese mathematicians – math110 Aug 13 '13 at 1:01
• Is there an organical way to come up with the one-liner solution? – chubakueno Aug 14 '13 at 2:37

Let $T$ be the function in question: $$T(a,b,c,d) = a b^2(b-c) + b c^2 (c-d) + c d^2 (d-a) + d a^2 (a - b).$$ We wish to show $T(a,b,c,d)\ge 0$ if $a,b,c,d$ are the sides of a quadrilateral. (Presumably, $a$ is the side opposite $c$ and $b$ is opposite $d$, but it actually doesn't matter to the proof.)

Terminology We introduce the following terminology: If $x_1,x_2,x_3,x_4$ are four real numbers (possibly negative), we say $x_1,x_2,x_3,x_4$ are "quadrilateral" if the following constraint holds: \begin{eqnarray} (*)\ \ \ \ x_1 + x_2 + x_3 + x_4 &\ge& 2 \max\{x_1,x_2,x_3,x_4\}. \end{eqnarray}

We also say that $x_1,x_2,x_3,x_4$ are "linear quadrilateral" if $(*)$ holds with equality.

Obviously, the sides $a,b,c,d$ of a quadrilateral are, well, quadrilateral. If the quadrilateral is degenerate so that its four vertices fall on a line, then its sides are linear quadrilateral.

Basic Idea The basic idea of the proof is to continuously "shrink" the sides $a,b,c,d$ of the quadrilateral by equal amounts $x$ until the quadrilateral collapses and all four of its vertices fall on a line. In other words, $(a-x,b-x,c-x,d-x)$ are linear quadrilateral. In step 3 below, we show this shrinking process decreases $T$, i.e., $T(a-x,b-x,c-x,d-x)$ is decreasing in $x\ge 0$ until $a-x,b-x,c-x,d-x$ are linear quadrilateral. The complexity comes when one realizes that during this shrinking process, one of the sides may collapse through a point and its length become negative (in which case it no longer makes sense to talk about $a-x,b-x,c-x,d-x$ being sides of a quadrilateral). Step 1 handles this "negative" case. Step 2 handles the more natural case in which no side becomes negative during the shrinking process.

Step 1 Suppose $a,b,c,d\ge 0$ are quadrilateral and one of $a,b,c,d$ vanishes. Then $T(a,b,c,d)\ge 0$.

Proof Assume without loss of generality that $a=0$. Then $$T(0,b,c,d) = b c^3 - b c^2 d + c d^3.$$ Observe \begin{eqnarray} c \ge d &\implies& b c^3 - b c^2 d \ge 0 \implies T(0,b,c,d) \ge 0 \\ d \ge b,c &\implies& c d^3 - b c^2 d \ge 0 \implies T(0,b,c,d) \ge 0. \end{eqnarray} The only other case not covered by these two conditions is $b> d > c$. In this case, we use use the fact that $a,b,c,d$ are quadrilateral to deduce $b\le c+d$. Since $c-d<0$, \begin{eqnarray} T(0,b,c,d) &=& b c^2(c-d) + c d^3 \\ &\ge& c^2 (c+d)(c-d) + c d^3 \\ &=& c^4 - c^2 d^2 + c d^3 \\ &=& c^4 + c d^2 (d - c) \\ &\ge& 0. \end{eqnarray} In any case, $T(0,b,c,d)\ge 0$.

Step 2 Suppose $a,b,c,d\ge 0$ are linear quadrilateral. Then $$T(a,b,c,d)\ge 0.$$

Proof Without loss of generality, suppose $d=\max\{a,b,c,d\}$, so $d=a+b+c$.

By direct computation, \begin{eqnarray} T(a,b,c,a+b+c) &=& a^4 - a^2 b^2 + a b^3 + a^3 c + a b^2 c + b^3 c + a^2 c^2 + 3 a b c^2 \\ & & \ + 2 b^2 c^2 + 2 a c^3 + 3 b c^3 + c^4. \end{eqnarray}

Luckily, the only summand that can possibly be negative is $-a^2 b^2$. Observe \begin{eqnarray} a\ge b &\implies& a^4 - a^2 b^2 \ge 0 \\ a\le b &\implies& a b^3 - a^2 b^2 \ge 0. \end{eqnarray} Thus, $$T(a,b,c,a+b+c) \ge a^4 + a b^3 - a^2 b^2 \ge 0.$$

Step 3 Suppose $a,b,c,d$ are linear quadrilateral and not all equal. Suppose the sum of any two of $a,b,c,d$ is non-negative. Then for $x\ge 0$, the mapping $$x \mapsto T(a+x,b+x,c+x,d+x)$$ is strictly increasing in $x$.

Proof

Without loss of generality, let $d=\max\{a,b,c,d\}$. Since $a,b,c,d$ are linear quadrilateral, $$d = a + b + c.$$

Direct computation shows $$T(a+x,b+x,c+x,d+x) = T(a,b,c,d) + A x + B x^2$$ where \begin{eqnarray} A &=& a^3 - a^2 b + 2 a b^2 + b^3 - 2 a b c - b^2 c + 2 b c^2 + c^3 \\ & & \ + 2 a^2 d - 2 a b d - 2 a c d - 2 b c d - c^2 d - a d^2 + 2 c d^2 + d^3. \end{eqnarray} and \begin{eqnarray} B &=& 2 a (a - b) + a (b - c) + 2 b (b - c) + b (c - d) \\ & & \ + 2 c (c - d) + (a - b) d + c (-a + d) + 2 d (-a + d) \\ &=& (a-c)^2 + (b-d)^2 + \frac{1}{2}(a-b)^2 + \frac{1}{2}(a-d)^2 + \frac{1}{2}(b-c)^2 + \frac{1}{2}(c-d)^2. \end{eqnarray} Clearly, $B>0$ because $a,b,c,d$ are not all the same.

Now substitute $d=a+b+c$ in the expression for $A$ and simplify: $$A = 3 a^3 + 2 a b^2 + 2 b^3 + 3 a^2 c + 2 b^2 c + 3 a c^2 + 6 b c^2 + 3 c^3.$$

Suppose $a<0$. Since the sum of any two of $a,b,c,d$ is non-negative, $b,c,d\ge |a|$. Observe \begin{eqnarray} A &=& 3 a^3 + 2 a b^2 + 2 b^3 + 3 a^2 c + 2 b^2 c + 3 a c^2 + 6 b c^2 + 3 c^3 \\ &=& (3 a^3 + 3a^2 c) + (2 a b^2 + 2 b^3) + (3 a c^2 + 3 c^3) + 2 b^2 c + 6 b c^2 \\ &\ge& 0. \end{eqnarray} (All the quantities in parentheses are non-negative.)

Suppose $b<0$. Because the sum of any two of $a,b,c,d$ is non-negative, $a,c,d\ge |b|$. We have \begin{eqnarray} A &=& 3 a^3 + 2 a b^2 + 2 b^3 + 3 a^2 c + 2 b^2 c + 3 a c^2 + 6 b c^2 + 3 c^3 \\ &=& (2 b^3 + 2 a b^2) + (6 b c^2 + 3 c^3 + 3 a c^2) + 3 a^3 + 3 a^2 c + 2 b^2 c \\ &\ge& 0. \end{eqnarray}

Finally, suppose $c<0$. Then $a,b,d\ge |c|$ and \begin{eqnarray} A &=& 3 a^3 + 2 a b^2 + 2 b^3 + 3 a^2 c + 2 b^2 c + 3 a c^2 + 6 b c^2 + 3 c^3 \\ &=& (3 a^2 c + 3 a^3) + (2 b^2 c + 2 a b^2)+ (3 c^3 + 3 a c^2) + 6 b c^2 + 2 b^3 \\ &\ge& 0. \end{eqnarray}

Since $B>0$ and $A\ge 0$, the result follows.

Step 4 Suppose $a,b,c,d\ge 0$ are quadrilateral and not all the same. Then $T(a,b,c,d) > 0$.

Proof Make the following definitions: \begin{eqnarray} x_0 &=& \frac{1}{2}(a+b+c+d - 2\max\{a,b,c,d\}) \\ A &=& a - x_0 \\ B &=& b - x_0 \\ C &=& c - x_0 \\ D &=& d - x_0. \end{eqnarray}

It is easy to see $A,B,C,D$ are linear quadrilateral. Furthermore, the sum of any two of $A,B,C,D$ is non-negative. For example, \begin{eqnarray} A+B &=& a + b - 2 x_0 \\ &=& 2\max\{a,b,c,d\} - c - d \\ &\ge& 0. \end{eqnarray} All the other cases are just as easy.

Consider the function $f:[0,\infty)\to\mathbb{R}$ defined by $$f(x) = T(A+x,B+x,C+x,D+x).$$ It follows from step 3 that $f$ is strictly increasing.

Since $a,b,c,d$ are quadrilateral, $x_0\ge 0$, so $T(a,b,c,d) = f(x_0) \ge f(0)$. If $A,B,C,D$ are all non-negative, then $f(0)\ge 0$ by step 2 and we are done.

Let $m=\min\{a,b,c,d\}$ and suppose one of $A,B,C,D$ is negative, so $m < x_0$. Then $T(a,b,c,d) = f(x_0) > f(m)$. But by step 1, $f(m)\ge 0$ and we are done.

• Your solution uses the same basic idea as mine, but don't you think it is easier to use AM-GM as I did to finish it after that? – user21820 Dec 8 '13 at 9:18
• @user21820 You only cover what I call "Step 2", which is just a couple lines of my proof. You assert that decreasing $(a,b,c,d)$ simultaneously by $t$ reduces the desired expression, but you don't prove this. (I do in my Steps 3 and 4.) Furthermore, note that when you perform the reduction, at least one of the sides can actually become negative. (Take $0,100,100,100$, for example. In this case, one reduces by 50, making the shortest side go negative.) One needs to think about that possibility. – Will Nelson Dec 8 '13 at 9:41
• @user21820 It's not anywhere close to obvious that the reduction causes $T(a,b,c,d)$ to decrease. It's a ton of algebra. I proved that in my Steps 3 and 4. If you believe it's trivial, you might want to stare at my Step 4, in particular, and see if there's much unwarranted complexity. I don't see it. Also, it's not a problem of a side going to $0$, it's a problem of a side going negative, as in the $0,100,100,100$ case, which reduces to $-50,50,50,50$. If you allow that, you'll need to prove results when one of the sides might be negative, which can obviously introduce a lot more trickiness. – Will Nelson Dec 8 '13 at 10:00
• You're right that I didn't check that the reduction works. As for the negative case, I thought we can decrease until either one side is zero or the quadrilateral shrinks to a line? – user21820 Dec 8 '13 at 10:01
• @user21820 I wouldn't delete it if I were you. You've got lots of upvotes. There are some holes (some large) in the answer, but it's good as far as it goes. I am beginning to wonder how closely and carefully people read answers here, though! – Will Nelson Dec 8 '13 at 23:00

WLOG $a = \max(a,b,c,d)$

Let $t = \frac{1}{2} (b+c+d-a) \ge 0$ because $a$,$b$,$c$,$d$ are sides of a quadrilateral

Then decreasing $(a,b,c,d)$ simultaneously by $t$ reduces the desired expression

And $a-t = (b-t)+(c-t)+(d-t)$

Thus it suffices to minimize the expression when $a=b+c+d$, which reduces to: \begin{align} &(b+c+d) b^2 (b-c) + b c^2 (c-d) - c d^2 (b+c) + d (b+c+d)^2 (c+d) \\ & = \left( b^4 - b^2 c^2 + b^3 d - b^2 c d \right) + \left( b c^3 - b c^2 d \right) - c d^2 (b+c) + d (b+c+d) (b+c+d) (c+d) \\ & \ge \left( b^4 - b^2 c^2 + b^3 d - b^2 c d \right) + \left( b c^3 - b c^2 d \right) - c d^2 (b+c) + \Big( d (b+c) d c + d (b+c) (b+c) c \Big) \\ & \ge \left( b^4 - b^2 c^2 + b^3 d - b^2 c d \right) + \left( b c^3 - b c^2 d \right) + b (b+c) c d \\ & \ge b^4 - b^2 c^2 + b c^3 \\ & \ge 0 \quad\text{because} \quad \frac{1}{3} ( b^4 + 2 b c^3 ) \ge b^2 c^2 \quad \text{by AM-GM} \end{align}

• So just to be clear: this answer asserts but does not prove that decreasing $a,b,c,d$ simultaneously reduces the desired expression. (This is non-trivial. It's step 3 of my proof.) Furthermore, this reduction can result in negative lengths, which adds a lot of complexity. See steps 1 and 4 of my answer. – Will Nelson Dec 8 '13 at 20:50

WLOG, assume that $$a = \max(a, b, c, d)$$. Let $$x = a- b, \ y = a-c, \ z = a-d$$. Then $$x, y, z \ge 0$$. Let $$w = b+c+d - a$$. Then $$w > 0$$ since $$a, b, c, d$$ are the sides of a quadrilateral.

The inequality is written as $$\frac{1}{4}Aw^2 + \frac{1}{4}Bw + \frac{1}{4}C \ge 0$$ where \begin{align} A &= 2\, x^2 - x\, y - 2\, x\, z + 2\, y^2 - y\, z + 2\, z^2,\\ B &= 2\, x^3 + 4\, x^2\, y - 2\, x\, y^2 - 4\, x\, y\, z + 2\, y^3 + 4\, y^2\, z - 2\, y\, z^2 + 2\, z^3,\\ C &= (x+3z)y^3 - 4xzy^2 + (3x^3+3x^2z-3xz^2+z^3)y. \end{align} It suffices to prove that $$A, B, C \ge 0$$. We have \begin{align} A = (x^2 - xy + y^2) + (y^2 - yz + z^2) + (z^2 - 2zx + x^2) \ge 0 \end{align} and \begin{align} B &= 2x^3 + 3x^2y + (x^2y - 2xy^2 + y^3) - 2(2xz)y + y^3 + 3y^2z + z^3 + (y^2z - 2yz^2 + z^3)\nonumber\\ &\ge 2x^3 + 3x^2y - 2(x^2+z^2)y + y^3 + 3y^2z + z^3\nonumber\\ &= 2x^3 + x^2y + y^3 + 2y^2z + (y^2z + z^3 - 2yz^2)\nonumber\\ &\ge 0. \end{align} And $$C\ge 0$$ follows from \begin{align} &4(x+3z)(3x^3+3x^2z-3xz^2+z^3) - (4zx)^2\nonumber\\ =\ & 12x^4+48x^3z+8x^2z^2-32xz^3+12z^4\nonumber\\ =\ & (12x^4 - 16x^3z + 24x^2z^2) + 64x^3z - 16x^2z^2 - 32xz^3 + 12z^4\\ \ge \ & 64x^3z - 16x^2z^2 - 32xz^3 + 12z^4 \nonumber \\ = \ & 4z(4x+3z)(2x-z)^2\nonumber\\ \ge \ &0. \end{align} We are done.

use the ptolemy inequality !

$ab^2(b-c)+bc^2(c-d)+cd^2(d-a)+da^2(a-b)\ge 0$$\Longrightarrow 2(ab^3+bc^3+cd^3+da^3)\ge$$b^2\cdot{a}(b+c)+c^2\cdot{b}(c+d)+d^2\cdot{c}(a+d)+a^2\cdot{d}(a+b)$$\Longrightarrow 2(ab^3+bc^3+cd^3+da^3)\ge$$b^2\cdot{a}l_{1}+c^2\cdot{b}l_{2}+d^2\cdot{c}l_{1}+a^2\cdot{d}l_{2}$$\Longrightarrow 8abcd\ge$$b^2\cdot{a}l_{1}+c^2\cdot{b}l_{2}+d^2\cdot{c}l_{1}+a^2\cdot{d}l_{2}$$\Longrightarrow 8\ge$$bl_{1}/cd+cl_{2}/ad+dl_{1}/ab+al_{2}/bc$

if $bl_{1}/cd,cl_{2}/ad,dl_{1}/ab,al_{2}/bc\ge2$

then, $l_{1}l_{2}abcd\ge bc\cdot(cd)^2+ad\cdot(ab)^2+ab\cdot(bc)^2+cd\cdot(ad)^2$

$\Longrightarrow$$ac+bd\ge 4\sqrt{abcd}$$\Longrightarrow$$l_{1}l_{2}\ge ac+bd$

which contradict with the ptolemy inequality !

Let $$\;$$ $$LHS=f(a,b,c,d)$$. Note that $$f(a,b,c,d)>f(a-k,b-k,c-k,d-k)$$ for $$0. So, WLOG we can take $$d=0$$ to prove the inequality. In that case we should show $$ab^2(b-c)+bc^3\ge0$$. If $$b\ge c$$ obviously we are done. For $$c\ge b \ge a$$ and $$c\ge a \ge b$$ cases arranging the inequality shows us $$ab^3+bc(c^2-ab)\ge0$$. Now, $$a\ge c \ge b$$ case remained. We will use the inequality $$a\ge c$$. Let's arrange the inequality again. Then, $$ab^2(b-c)+bc^3\ge cb^2(b-c)+bc^3=b^3c-b^2c^2+bc^3\ge0$$ $$\;$$ (by AM-GM $$b^3c+bc^3\ge 2b^2c^2$$ )