It is very similar to 2016 China Western MO P1

Do you have 2011-2016 Silk Road MO Problems?Thanks.

Let the $42$ numbers be $x_1\leq x_2\leq\cdots \leq x_{42}$. If for all $k=1,2,\ldots,20$ we have $x_{2k+1}x_{2k+2}>4x_{2k-1}x_{2k}$ then $$x_{41}x_{42}>4x_{39}x_{40}>\cdots>4^{20}x_1x_2 > 10^{12}$$which is impossible. Hence there exists a $k$ such that $x_{2k+1}x_{2k+2}\leq 4x_{2k-1}x_{2k}$. We choose the four numbers $w=x_{2k-1},x=x_{2k},y=x_{2k+1},z=x_{2k+2}$. By rearrangement, $zy+wx\geq zx+wy\geq zw+xy$ so it suffices to show that $$25(zx+wy)(zw+xy)\geq 16(zy+wx)^2.$$ Since $wx\leq zy\leq 4wx$, we have $(4wx-zy)(4zy-wx)\geq 0$, which is equivalent to $25wxyz\geq 4(zy+wx)^2$. Therefore, by AM-GM, $$25(zx+wy)(zw+xy)\geq 100 wxyz\geq 16(zy+wx)^2$$as required.

guangzhou-2015 wrote: It is very similar to 2016 China Western MO P1 2016 China Western MO P1 Let $a,b,c,d$ be real numbers such that $abcd>0$. Prove that:There exists a permutation $x,y,z,w$ of $a,b,c,d$ such that $$2(xy+zw)^2>(x^2+y^2)(z^2+w^2)$$