# Bounding the product of a quadrilateral's side lengths in terms of the lengths of its diagonals

Suppose we've a convex quadrilateral both of whose diagonals have length 2. Is it true that the product of the lengths of the quadrilateral's sides must be less than or equal to 5?

If we require, in addition, that the diagonals of the quadrilateral are perpendicular, this becomes equivalent to the inequality of this question.

A product of 5 is achieved when the diagonals are perpendicular and 3 of the vertices are colinear. I am also curious whether a similar configuration is optimal in the case where the diagonals have different lengths x and y. If so, we would obtain a bound on the product in terms of x and y by computing the product for this configuration.

Added: Here is an applet demonstrating the inequality. http://tube.geogebra.org/student/m905097

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+1 for the nice applet. – TonyK Mar 25 '15 at 21:42

This was fun.

We will in fact prove a stronger statement, that the sum of the products of opposite sides of such a quadrilateral is no more than $2\sqrt{5}$. The result then follows from an application of $AM-GM$. We will work with the algebraic analogue of this problem:

Let $a, b, c, d, x$ be positive reals with $a+b=c+d=2$ and $0\le x\le 1$. Prove that: $$\sqrt{(a^2+b^2+2abx)(c^2+d^2+2cdx)}+\sqrt{(a^2+d^2-2adx)(b^2+c^2-2bcx)}\le 2\sqrt{5}$$

(Note: here, $x$ is taken to be the nonnegative cosine of the angle between the two diagonal, WLOG supposing that the angle between the segments with length $a$ and $b$ is acute.)

First, we manipulate the expressions under the radicals, call them $L$ and $R$: \begin{align*} L&=(a^2+b^2+2abx)(c^2+d^2+2cdx) \\ &=(a^2+b^2)(c^2+d^2)+2x(ab(c^2+d^2)+cd(a^2+b^2))+4x^2abcd \\ &=(ac-bd)^2+(ad+bc)^2+2x(ad+bc)(ac+bd)+x^2[(ac+bd)^2-(ac-bd)^2] \\ &=[x(ac+bd)+(ad+bc)]^2+(1-x^2)(ac-bd)^2 \end{align*} In a similar manner, we see that: $$R=[x(ac+bd)-(ab+cd)]^2+(1-x^2)(ac-bd)^2$$ Now let $p=x(ac+bd)+\frac{(ad+bc)-(ab+cd)}{2}$ and $q=|ac-bd|\sqrt{1-x^2}$. Using the fact that $\frac{ad+bc+ab+cd}{2}=\frac{(a+c)(b+d)}{2}=2$, our inequality can be rewritten as: $$\sqrt{(p+2)^2+q^2}+\sqrt{(p-2)^2+q^2}\le 2\sqrt{5}$$ We will now prove the key lemma:

Lemma: $p^2+5q^2\le 5$

Proof: Using the fact that $ad+bc-(ab+cd)=(a-c)(d-b)=4(a-1)(b-1)$, this inequality can be rewritten as: $$5[a(2-a)-b(2-b)]^2(1-x^2)+[x(a(2-a)+b(2-b))+2(a-1)(b-1)]^2\le 5$$ Upon the translation $a\Rightarrow 1+y, b\Rightarrow 1+z$ for $y, z\in [-1, 1]$, this simplifies to: y, z\ge 9$$5(y^2-z^2)^2(1-x^2)+[x(2-y^2-z^2)^2-2yz]^2\le 5$$ $$\Leftrightarrow 4x^2(1+3y^2z^2-y^2-y^4-z^2-z^4)+4xyz(2-y^2-z^2)+5y^4+5z^4-6y^2z^2\le 5$$

Let this expression be $F(x, y, z)$. From this formulation, we see that since $F(x, y, z)\le F(x, |y|, |z|)$, we can say $y, z\ge 0$ WLOG. Now, we take two cases:

Case 1: $x\le \frac{3}{4}$. In this case, note that: \begin{align*}F(x, 1, yz)-F(x, y, z)&= (5-4x^2)(1-y^4)(1-z^4)-(4x^2+4xyz)(1-y^2)(1-z^2) \\ &=(1-y^2)(1-z^2)[(5-4x^2)(1+y^2)(1+z^2)-4x^2-4xyz] \\ &\ge (1-y^2)(1-z^2)[(5-\frac{9}{4})(1+y^2)(1+z^2)-\frac{9}{4}-3yz] \\ &= (1-y^2)(1-z^2)[\frac{1}{2}+\frac{11}{4}(y-z)^2+\frac{5}{2}yz+\frac{11}{4}y^2z^2] \\ &\ge 0 \end{align*}
It follows that it suffices to prove the inequality for $z=1$. But we see that: \begin{align*}F(x, y, 1)&=4x^2(-y^4+2y^2-1)+4xy(1-y^2)+5y^4-6y^2+5 \\ &=5-5y^2(1-y^2)-[2x(1-y^2)-y]^2 \\ &\le 5 \end{align*} And so this case is complete.

Case 2: $x>\frac{3}{4}$. In this case, note that: $$4xyz(2-y^2-z^2)\le 4xyz(2-2yz)=8xyz(1-yz)\le (4x^2+4)yz(1-yz)$$ Therefore, it suffices to show that: $$4x^2(yz(1-yz)+3y^2z^2+1-y^4-z^4-y^2-z^2)+4yz(1-yz)+5y^4-6y^2z^2+5z^4\le 5$$ This last function is linear in $x^2$, so its maximum occurs at the endpoints of its domain. Therefore, it suffices to check the inequality for $x=\frac{3}{4}$ and $x=1$. The latter gives: $$y^4+z^4-2y^2z^2-y^2-z^2+2yz+4\le 5$$ $$\Leftrightarrow (y^2-z^2)^2-(y-z)^2\le 1$$ which is true because both terms are in $[0, 1]$. Finally, for $x=\frac{3}{4}$ the inequality becomes: $$\frac{11}{4}y^4+\frac{11}{4}z^4-\frac{11}{2}y^2z^2-\frac{9}{4}(y^2+z^2-yz)+\frac{9}{4}\le 5$$ $$\Leftrightarrow \frac{11}{4}(y^2-z^2)^2-\frac{9}{4}(y^2+z^2-yz)\le \frac{11}{4}$$ Which is true because $y^2+z^2-yz\ge 0$.

Now, back to the main inequality. The lemma can be rewritten as: $$(p^2+q^2+4)^2-16p^2\le (6-p^2-q^2)^2$$ $$\Leftrightarrow [(p+2)^2+q^2][(p-2)^2+q^2]\le (6-p^2-q^2)^2$$ Because $p^2+q^2<5<6$, this implies that: $$\sqrt{[(p+2)^2+q^2][(p-2)^2+q^2]}\le 6-p^2-q^2$$ $$\Leftrightarrow [\sqrt{(p+2)^2+q^2}+\sqrt{(p-2)^2+q^2}]^2\le 20$$ Finally, taking the square root of this gives the desired result. $\Box$

Equality holds when $p=0, q=1$, which corresponds to $(a, b, c, d)=(2, 1, 0, 1)$ and analogous tuples.

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Thanks for the reply! I will make some coffee and then read this :) – Mike F Mar 28 '15 at 17:06
Hopefully there are no errors :) I had a great time trying this one, it's a very nice inequality in my opinion. – Apple Mar 28 '15 at 17:25
Hey I've gone ahead and awarded the bounty, although I am wondering whether your proof can be simplified somewhat. Small correction: in your 3rd paragraph it looks like you meant $a+c=b+d=2$. – Mike F Mar 30 '15 at 18:38