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

Question

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 $(xy+zw)^2$.

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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?

Answer 1

$$36=x^2+y^2-\frac{xy}{2}=\left(x-\frac{y}{4}\right)^2+\left(\frac{\sqrt{15}y}{4}\right)^2$$ 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.

Answer 2

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.

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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).

Answer 3

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.

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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.