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Before I show my method, I'll post an example question. (Also, this one is the start of my wonder.)

$x^2+y^2=1, z^2+w^2=1$. Then, find the minimum and maximum of $xw+yz$.

Of course, It can be easily solved with the Brahmagupta-Fibonacci Inequality, but I'll try to use my strategy.

\begin{align} & \alpha=\arcsin(x), \beta=\arcsin(z). \\ & \therefore \arccos(y)=\alpha, \arccos(w)=\beta. \\ & xw+yz=\sin\alpha\cos\beta+\cos\alpha\sin\beta=\sin(\alpha+\beta). \\ & \therefore \min(xw+yz)=-1, \max(xw+yz)=1. \end{align}

Like this, I found that if the variables satisfy $x^2+y^2=1$, we can use it as a trigonometric function. (Maybe a lot of people know about this.)

But, my question is:

Is this method always valid when the variables $x, y$ satisfies $x^2+y^2=1$, and the question is asking about the minimum, maximum, or the range of quadratic polynomials?

If not, please let me know the counterexample of this.

RDK
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    Your question is rather vague. What does it mean to say that the question is all about quadratic polynomials? Can you formulate a logically precise version of your question? – Lee Mosher Jul 03 '22 at 14:56
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    If $x^2+y^2=1$, there exists $a$ such that $x=\cos a,y=\sin a$. This is true even if we allow $x,y,a\in\mathbb C$. – eyeballfrog Jul 03 '22 at 15:15
  • Further, if $x,y$ satisfy some quadratic equation that does not define a parabola, then there either exist constants $a,b,c,d,\theta$ such that $x=a\cos\theta+b,y=c\sin\theta+d$, or the same but with hyperbolic functions. Finding these constants is nontrivial, but they will exist. – eyeballfrog Jul 03 '22 at 16:04
  • Umm... just for the one who has downvoted this post, please tell me why you downvoted this post so that I can edit this post. – RDK Jul 03 '22 at 17:37

1 Answers1

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Use Cauchy–Schwarz inequality: $\vert \langle (x,y),(w,z)\rangle\vert \leq \Vert (x,y)\Vert \cdot \Vert (w,z)\Vert$. Equivalently, $$ \vert xw+yz\vert \leq \sqrt{x^2+y^2}\cdot \sqrt{w^2+z^2} =1$$

Senna
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  • Yes, this is also a nice solution, but this was just an example to show my method. In fact, this has a lot of solutions. For example, $(x^2+y^2)(z^2+w^2)=(xw+yz)^2+(xz-yw)^2$, so $(xw+yz)^2$ has a maximum of $1$, when $xz=yw$. – RDK Jul 03 '22 at 17:33