Interested to see a solution

StarT?

Clearly the desired inequality is just proving that there are 4 numbers $a,b,c,d$ st : $$(\frac{ac}{bd} +1)^2 \geq 9(\frac{a}{b} - \frac{c}{d})^2$$$*$ Now let for any two numbers $a,b$ , $f(a,b)=min{\frac{a}{b} , \frac{b}{a}}$. So if $a,b$ have the same sign , $f(a,b)$ would be a number between $0,1$. We have some cases that are pretty similar to each other : Case 1. All 9 numbers are non-negative , that is : $0 \leq a_1<...<a_9$ Consider the 4 numbers $f(a_1,a_2),f(a_3,a_4),f(a_5,a_6),f(a_7,a_8)$ . So two of them lie in one of three intervals $[0,1/3),[1/3,2/3),[2/3,1)$. For example $f(a,b)-f(c,d) \leq 1/3$ and consider a,b,c,d . So clearly $\frac{ac}{bd}$ is positive and the LHS of * would be larger than 1 and RHS less than 1 Case 2.one number is negative. Let $a_9$ be negative. But it doesn't really matter since in the pervious case we just used that $a_1,... a_8$ are non-negative Case 3,4,5. Two and three and four numbers are negative respectively. Again the same as above but consider the $f$ of those two negatives and clearly their product would be positive and we can do the same as case 1. In other case we can just change the sign of our 9 numbers and we'll arrive to this recent cases. So in all case the statement is true and we're done

Mehrshad wrote: Prove that among any $9$ distinct real numbers, there exist $4$ distinct numbers $a,b,c,d$ such that $$(ac+bd)^2\ge\frac{9}{10}(a^2+b^2)(c^2+d^2)$$ Who is the author?

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Interesting.

Take $a,b$ arbitrary and assume that $(ac+bd)^2<\frac{9}{10}(a^2+b^2)(c^2+d^2)$ for the 21 unorder pairs taken from the $7$ remaining numbers. Write $\textbf{x}=a\hat{i}+b\hat{j}$ and $\textbf{y}=c\hat{i}+d\hat{j}$. The inequality from above translates to $(\textbf{x} \cdot \textbf{y})^{2}<\frac{9}{10}(\left|\textbf{x} \right|\left|\textbf{y} \right|)^2$, which translates to $\left| \angle (\textbf{x}, \textbf{y}) \right|>arcos(\frac{3}{\sqrt{10}})>0.3$. Now, since this is true for any of the $21$ vectors $ \textbf{y}$ , we have $6.3>2\pi \geq 21 \cdot arcos(\frac{3}{\sqrt{10}})>21\cdot 0.3=6.3$. Contradiction!

Proposed by Navid Safaei.