7
$\begingroup$

If $a$, $b$, $c$ and $d$ are the four sides of a quadrilateral, the problem is to show that $ab^2(b-c)+bc^2(c-d)+cd^2(d-a)+da^2(a-b)\ge 0$. I've verified it to be true for quite a large number of values, but can't seem to come up with a proof for it. Does anyone have any ideas, perhaps some inequality results that can be applied to prove it?

$\endgroup$
  • 1
    $\begingroup$ Where does this inequality come from? $\endgroup$ – Stefan Kohl Dec 5 '13 at 17:46
  • 1
    $\begingroup$ Well, the inspiration for this inequality comes from an old IMO problem which asks to prove the following, for a triangle:- $a^{2}b(a - b) + b^{2}c(b - c) +c^{2}a(c - a)\ge 0.$ It is relatively easier to prove the latter, by substitution. $\endgroup$ – Train Heartnet Dec 5 '13 at 18:49
  • $\begingroup$ It's a nice question, but not research mathematics, IMHO $\endgroup$ – Igor Rivin Jan 13 '15 at 16:55
  • $\begingroup$ @IgorRivin: What do you think are the benefits of closing a year-old answered question like this? $\endgroup$ – Stefan Kohl Jan 13 '15 at 17:58
  • 1
    $\begingroup$ @StefanKohl I did not realize it was a year old... $\endgroup$ – Igor Rivin Jan 13 '15 at 19:40
3
$\begingroup$

Assuming the contrary, we get that the minimal value of $$ f(a,b,c,d)=ab^2(b−c)+bc^2(c−d)+cd^2(d−a)+da^2(a−b) $$ on the set $$ M=\{(a,b,c,d)\colon a,b,c,d\geq 0, a+b+c+d\leq 1, a+b+c\geq d,\dots\} $$ is negative. By the homogeneity, this value is achieved on $a+b+c+d=1$.

Later we will check that $f$ is nonnegative on the boundary faces $a=0$ and $a+b+c=d$. Thus the minimum is achieved at an interior point of the face defined by $a+b+c+d=1$. By the (Karush--)Kuhn--Tucker theorem, this point satisfies the system $$ -\lambda=b^2(b-c)-cd^2+3da^2-2dab=\cdots $$ for some nonnegative $\lambda$ (the other equalities are obtained by the cyclic permutation of variables). In particular, this means that $$ b^3+3da^2\leq b^2c+cd^2+2dab, \;\dots, $$ which sums up to $$ \sum a^3+2\sum ab^2\leq \sum a^2b+2\sum abc $$ (the sums are also over four cyclic permutations). On the other hand, by AM--GM we have $$ a^3+ab^2+bc^2+bc^2\geq 4abc, \quad a^3+ab^2\geq 2a^2b. $$ Summing up all these we get the converse; thus all these inequalities should come to equalities, and hence $a=b=c=d$; but $f(a,a,a,a)=0$. A contradiction.

It remains to check the inequality on the other boundary faces. If $a=0$ then the inequality looks like $bc^2(d-c)\leq cd^3$; clearly we may assume that $d\geq c$, otherwise the inequality is trivial. Since $b\leq c+d$, we have $ bc^2(d-c)\leq c^2(d^2-c^2)\leq cd\cdot d^2$, as required.

Finally, if $a=b+c+d$ then we may substitute this into our function getting (here Maple was used...) the large sum with many positive terms (including $b^4+bc^3$) and only one negative term $-b^2c^2$. But it is easy to see that $b^4-b^2c^2+bc^3\geq 0$ for all nonnegative $b$ and $c$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Seems to be directly plagiarized from this answer $\endgroup$ – user574848 Mar 24 at 11:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.