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For any nine distinct real numbers,there exsit four distinct real number $x,y,z,w$,such $$|xz+yw|\ge |\sqrt{5}(xw-yz)|$$

I think can use pigeonhole principle to solve it?Thanks

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    Potentially helpful $$(xz+yw)^2+(xw-yz)^2=(x^2+y^2)(z^2+w^2)$$ – 2017-02-17
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    If, as it seems, the four $w,x,y,z$ are to be chosen from among the nine initially assumed distinct real numbers, there seems no need to say the latter are distinct. – 2017-02-17
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    I must be missing something, but looks like it is enough to prove it for integers (which I expect will be easier, haven't tried). Once we know it is true for integers, it is true for rationals, and since rationals are dense and every infinite sequence taking values from $\binom{9}{4}$ symbols has some symbol that repeats infinitely often... – 2017-02-17
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    There is a somewhat similar problem (with solution) at https://www.math.ust.hk/excalibur/v1_n1.pdf – among any nine distinct real numbers, there are two, $a$ and $b$, such that $0<(a-b)/(1+ab)<\sqrt2-1$. Note also that $xz+yw$ is the imaginary, and $xw-yz$ the real, part of $(x+iy)(w+iz)$. – 2017-02-22

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It seems that actually just $7$ real numbers suffice, though I have had to use a calculator, so some clever hand written argument may suffice for $9$ real numbers.

First lets consider when $|xy+zw|\ge|\sqrt{5}(xw-yz)|$. I apologise for the lack of diagram - you may wish to draw one as you read.

We have $xy+zw=\binom{x}{z}\cdot\binom{y}{w}=|A||B|\mathrm{cos}(\theta)$ where $\theta$ is the angle between $A=\binom{x}{z}$ and $B=\binom{y}{w}$ and $xw-yz=\binom{x}{z}\cdot\binom{w}{-y}$. Notice that $\binom{w}{-y}$ can be obtained by rotating $\binom{y}{w}$ $90^\circ$ clockwise (it turns out in this question that degrees will be easier than radians). Therefore $|\binom{x}{z}\cdot\binom{w}{-y}|=|A||B||\mathrm{cos}(90^\circ-\theta)|=|A||B||\mathrm{sin}(\theta)|$. Putting this together $|xw-yz|/|xy+zw|=|\mathrm{sin}(\theta)/\mathrm{cos}(\theta)|=|\mathrm{tan}(\theta)|$ so the question is instead when is $|\mathrm{tan}(\theta)|\le\frac{1}{\sqrt{5}}$.

I used a calculator, but there may be a clever hand written calculation you could use instead. $\mathrm{tan}^{-1}(\frac{1}{\sqrt{5}})\approx 24.09^\circ$. In particular, if $|\theta|\le 24^\circ$ or $|180^\circ-\theta|\le 24^\circ$ then the inequality holds.

Now, as you suspected we use the pigeonhole principle.

From the $7$ real numbers, choose three disjoint pairs $(x_1,x_2),(x_3,x_4),(x_5,x_6)$ such that each number from a given pair has the same sign - it is easy to show that this is possible! In particular each $\binom{x_i}{x_{i+1}}$ is in either the upper right or lower left quadrant of the real plane - note also that $x_i\ne x_{i+1}$ so they do not lie on the line $x=y$.

Now we split the union of these quadrants (minus the line $x=y$) into $4$ sets. $$A=\left\{\binom{x}{y}|x=0\hspace{0.2cm}or\hspace{0.2cm}\mathrm{tan}^{-1}(y/x)>77.5^\circ\right\}$$ $$B=\left\{\binom{x}{y}|45^\circ<\mathrm{tan}^{-1}(y/x)\le77.5^\circ\right\}$$ $$C=\left\{\binom{x}{y}|22.5^\circ\le \mathrm{tan}^{-1}(y/x)<45^\circ\right\}$$ $$D=\left\{\binom{x}{y}|0^\circ\le\mathrm{tan}^{-1}(y/x)<22.5^\circ\right\}$$

Notice that any two points lying in one of these for sets create an angle at the origin $\theta$ with $|\theta|\le 22.5^\circ$ or $|180^\circ-\theta|\le 22.5^\circ$. Therefore, to find $4$ real numbers which satisfy the inequality it suffices to use $4$ to define two points in one of these sets.

From each pair place the point $\binom{x_i}{x_{i+1}}$ on the plane. By the pidgeonhole principle, we must have two points lie in either $A\cup D$ or $B\cup C$. If a point $\binom{x}{y}$ lies in $C$ or $D$ then $\binom{y}{x}$ lies in $B$ or $A$ respectively. This means that by swapping $x_i$ with $x_{i+1}$ if necessary we obtain two points in $A$ or two points in $B$.