any solutions?the problem is easy......

sqing wrote: 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 $$2(xy+zw)^2>(x^2+y^2)(z^2+w^2)$$. any orther problems,I think CWMI have a good geometry problem.

Let no. Hence, $(a^2+b^2)(b^2+c^2)+(a^2+c^2)(b^2+d^2)+(a^2+d^2)(b^2+c^2)\geq2(ab+cd)^2+2(ac+bd)^2+2(ad+bc)^2$, which is $abcd\leq0$. Contradiction.

arqady wrote: Let no. Hence, $(a^2+b^2)(b^2+c^2)+(a^2+c^2)(b^2+d^2)+(a^2+d^2)(b^2+c^2)\geq2(ab+cd)^2+2(ac+bd)^2+2(ad+bc)^2$, which is $abcd\leq0$. Contradiction. what do you mean?

I meant that we got a contradiction.

Suppose otherwise. Sum the three inequalities \begin{align*} (x^2+y^2)(z^2+w^2) &\geq 2(wz+xy)^2 \\ (x^2+z^2)(y^2+w^2) &\geq 2(xz+wy)^2 \\ (w^2+x^2)(z^2+y^2) &\geq 2(wx+yz)^2. \end{align*}This yields $6wxyz \leq 0$, impossible.

@ HamstPan38825 by summing the right side of the inequalities and deleting the same monomials, shouldn't we have: 12xyzw<=0

Click to reveal hidden text