# Find maximum of $(xy+zw)^2$ subject to $xz+yw=30$ etc.

If for $$x,y,z$$ positive reals, $$x^2+y^2-\frac{x y}{2}=w^2+z^2+\frac{wz}{2}=36$$ and $$xz+yw=30$$ find maximum value of $$(xy+zw)^2$$.

By Cauchy-Schwarz Inequality, we get $$(x^2+w^2)(y^2+z^2) \geq (xy+zw)^2$$ The so by the equality case of cauchy Schwartz Inequality, we get maximum value of $$(xy+zw)^2$$ will be achieved when $$\frac{x}{y}=\frac{w}{z}=k$$ for some non zero real $$k$$. Using this we get $$x=yk$$ and $$w=zk$$ but for this, we get some really ugly equations that I'm unable to solve. Is there a better way out?

• I guess that we can use geometry with the law of cosines, but still don't know how.
– NN2
May 25 at 22:19
• Note that with your approach, because $x^2 + w^2$ and $y^2 + z^2$ are not fixed values, you can't say much about the actual maximum. What you know is that there is an upper bound of $\max ( x^2 + w^2) (y^2 + z^2)$, but I doubt that can be achieved with the given conditions. $\quad$ EG If you look at my solution, what $x/y = w/z = k$ means is that these are on the apollonius circles of that ratio $k$. However, we then can't get $ABCD$ to be a cyclic quad unless $k = 1$, which it isn't. May 25 at 22:51

$$36=x^2+y^2-\frac{xy}{2}=\left(x-\frac{y}{4}\right)^2+\left(\frac{\sqrt{15}y}{4}\right)^2$$ and $$36=w^2+z^2-\frac{wz}{2}=\left(w+\frac{z}{4}\right)^2+\left(\frac{\sqrt{15}z}{4}\right)^2$$ Thus, there are $$\alpha$$ and $$\beta$$, for which $$x-\frac{y}{4}=6\cos\alpha,$$ $$\frac{\sqrt{15}y}{4}=6\sin\alpha,$$ $$w+\frac{z}{4}=6\cos\beta$$ and $$\frac{\sqrt{15}z}{4}=6\sin\beta.$$ Thus, $$xz+wy=30$$ gives $$\sin(\alpha+\beta)=\frac{5\sqrt{15}}{24}$$ and by C-S $$xy+zw=30\left(\frac{1}{\sqrt{15}}\sin(\alpha-\beta)+\cos(\alpha-\beta)\right)\leq$$ $$\leq30\sqrt{\left(\frac{1}{15}+1\right)(\sin^2(\alpha-\beta)+\cos^2(\alpha-\beta))}=\sqrt{960},$$ which gives $$(xy+wz)^2\leq960.$$ The equality occurs for $$\left(\frac{1}{\sqrt{15}},1\right)||(\sin(\alpha-\beta),\cos(\alpha-\beta)),$$ which says that $$960$$ is a maximal value.

• This seems almost magical that everything cancels out nicely. Is there a motivating factor? $\quad$ FWIW, equality occurs at $\alpha - \beta = \sin^{-1} \frac{1}{4}$. May 25 at 23:12
• @Calvin Lin If $p^2+q^2=1$, so we can use a trigonometric substitution. It's really interesting that during a work with $xz+wy$ we found out that $\sin\alpha\sin\beta$ disappears. May 26 at 5:02

Hint: Consider a cyclic quadrilateral $$ABCD$$ with $$AB = x, BC = y, CD = z, DA = w$$ and $$\angle ABC = \cos^{-1} \frac{1}{4}$$.

Steps towards a solution. Find in any minor gaps as needed. If you're stuck, explain what you've tried.

• Verify that the conditions imply such a setup is possible. In particular, $$AC = 6$$.
• Ptolemy's theorem gives us $$AC \cdot BD = xz + yw = 30$$.
• By considering area $$[ABCD]$$, show that $$\frac{1}{2} (AB \cdot BC + CD \cdot DA) \sin ( \angle ABC) = [ ABCD] \leq \frac{1}{2} AC \cdot BD = 15.$$
• Hence $$xy+zw \leq \frac{ 30 } { \frac{ \sqrt{15}} { 4} }$$
• Thus $$(xy + zw)^2 \leq 960$$.

Currently, this is merely an upper bound. Let's state all of the conditions for equality:

• $$ABCD$$ is a cyclic quad.
• $$\angle ABC = \cos^{-1} \frac{1}{4}$$.
• $$AC = 6$$.
• $$BD = 5$$.
• $$AC$$ and $$BD$$ are perpendicular.

Let's show that such a cyclic quad exists:

• Start with $$AC=6$$,
• Draw the locus of the circle with $$\angle ABC = \cos^{-1}\frac{1}{4}$$ , $$\angle ADC = \pi - \cos^{-1} \frac{1}{4}$$.
• Find the diameter $$d= \frac{6} { \sin \angle ABC} = 8 \sqrt{ \frac{3}{5} }$$
• Show that there is a perpendicular chord of length 5 that intersects $$AC$$ because $$\sqrt{d^2 - 6^2} < 5 < d$$.
• This gives us $$BD$$, and hence such a cyclic quad exists.

Hence, equality can be achieved and the maximum is 960.

Notes

1. I can't be bothered to find the exact values of $$x, y, z, w$$.
• River Li found the values, see comment below.
2. This approach will not work if $$BD < \sqrt{ D^2 - 6^2 }$$ or $$BX > D$$ as such a cyclic quad doesn't exist.
• Similarly in Michael's approach, that results in $$| \sin (\alpha + \beta) | > 1$$, though there might be some hope of salvaging from the algebra.
3. [River Li indicates this is incorrect, let me review... ] Per my comment below, if we set $$AB = x, BC = y, CD = \color{red}{w}, DA = \color{red}{z}$$ and $$\angle ABC = \cos^{-1} \frac{1}{4}$$, then we get $$\frac{1}{2} (xz + yw) \sin (\angle ABC) = [ABCD] \leq \frac{1}{2} ( xy + zw)$$, which gives us the minimum $$(xy+zw)^2 \geq 843 \frac{3}{4}$$ (again, verify that equality can hold).
• Several times, I found that for some inequality problems, geometric approach is powerful. Sometimes, pure algebraic or pure analysis solutions are very difficult to find. May 26 at 1:58
• FYI, the optimal $x, y, z, w$ is given by $x = \frac15\,\sqrt {480+15\,\sqrt {335}+25\,\sqrt {15}}$, $y = \frac15\,\sqrt {480-15\,\sqrt {335}+25\,\sqrt {15}}$, $z = \frac15\,\sqrt {480-15\,\sqrt {335}-25\,\sqrt {15}}$, $w = \frac15\,\sqrt {480+15\,\sqrt {335}-25\,\sqrt {15}}$. May 26 at 3:36
• @NN2 I also have a pure algebraic solution. I think geometric solution is natural. May 26 at 14:55
• @NN2 Mainly experience. You were already saying "use geometry with law of cosines", and so we have the 2 triangles with the same base length, so it's natural to put them together and see what happens, which is that we get a cyclic quad. That's a great picture, so how do the other conditions relate? Depending on how we set up the lengths, $xz+yw$ could be the product of consecutive sides (which then relate to area) or the product of opposite sides (which is Ptolemy's theorem). .... May 26 at 15:01
• Think about what $xy+zw$ relates to then. In the former, we get product of opposite sides (Ptolemy's), and in the latter, we get product of consecutive sides (area). Putting that together, the latter case gives us a maximum bound via [ABCD] (and the former case gives us a minimum bound -> TRY THIS!) May 26 at 15:02

Remarks: Michael Rozenberg and Calvin Lin gave very nice solutions. Here I give a pure algebraic solution. We use computer to motivate the solution. We have the following relation: \begin{align*} \left. \begin{array}{r} x, y, z \in \mathbb{R}\\ x^2 + y^2 - xy/2 = 36\\ w^2 + z^2 + wz/2 = 36\\ xz + yw = 30 \end{array} \right\} \Longrightarrow (xy + zw)^2 + \frac{5}{67}(5xy - 5zw - 48)^2 = 960. \end{align*} This enables us to write down the following solution.

Problem. Let $$x, y, z, w$$ be real numbers such that \begin{align*} x^2 + y^2 - xy/2 &= 36, \tag{1}\\ w^2 + z^2 + wz/2 &= 36, \tag{2}\\ xz + yw &= 30. \tag{3} \end{align*} Find the maximum of $$(xy + zw)^2$$.

Answer: The maximum of $$(xy + zw)^2$$ is $$960$$ when e.g. \begin{align*} x &= \frac15\,\sqrt {480+15\,\sqrt {335}+25\,\sqrt {15}}, \\ y &= \frac15\,\sqrt {480-15\,\sqrt {335}+25\,\sqrt {15}}, \\ z &= \frac15\,\sqrt {480-15\,\sqrt {335}-25\,\sqrt {15}}, \\ w &= \frac15\,\sqrt {480+15\,\sqrt {335}-25\,\sqrt {15}}. \end{align*}

Proof.

It suffices to prove that $$960 - (xy + zw)^2 \ge 0. \tag{4}$$

If $$xy = 0$$, from (2), we have $$\frac{3}{4}(w^2 + z^2) + \frac{(w + z)^2}{4} = 36$$ which results in $$\frac{3}{4}(w^2 + z^2) \le 36$$ and $$w^2 + z^2 \le 48$$. Thus, we have $$z^2w^2 \le \frac{(w^2 + z^2)^2}{4} = 576 < 960$$. (4) is true.

If $$zw = 0$$, from (1), we have $$\frac34(x^2 + y^2) + \frac{(x - y)^2}{4} = 36$$ which results in $$\frac34(x^2 + y^2) \le 36$$ and $$x^2 + y^2 \le 48$$. Thus, $$x^2y^2 \le \frac{(x^2 + y^2)^2}{4} = 576 < 960$$. (4) is true.

In the following, assume that $$xy \ne 0$$ and $$zw \ne 0$$.

From (3), we have $$x = \frac{30 - yw}{z}. \tag{5}$$

Using (5), $$[2y^2 \times (2) - 2z^2\times (1)]$$ gives $$120wy - 72y^2 + 30yz + 72z^2 - 1800 = 0$$ which results in $$w = \frac{12y^2 - 5yz - 12z^2 + 300}{20y}. \tag{6}$$

Using (5) and (6), (1) gives $$48y^4 + 29y^2z^2 + 48z^4 - 2400y^2 - 2400z^2 + 30000 = 0. \tag{7}$$

Using (7), we have \begin{align*} &(xy + zw)^2 + \frac{5}{67}(5xy - 5zw - 48)^2 - 960\\[6pt] ={}& \frac{(48y^4 + 29y^2z^2 + 48z^4 - 2400y^2 - 2400z^2 + 30000)(12y^2 - 5yz - 12z^2)}{6700y^2z^2}\\ ={}& 0 \end{align*} which results in $$(xy + zw)^2 \le 960.$$

We are done.