Not a difficult problem! The answer is sqrt(3). To prove it we can simply assume the contradictory. Using the fact that these two equality have a solution in R, assume them out as x1, x2 to get another two quadratic ones, like x^2+A1x+B1. Then by the assumption, A1^2>B1. Use p/q and q/p,r/s and s/r to establish two equations, and the problem will be down by Cauchy. The inequality holds when p=q=r=s=1.

dzy47 wrote: Not a difficult problem! The answer is sqrt(3). To prove it we can simply assume the contradictory. Using the fact that these two equality have a solution in R, assume them out as x1, x2 to get another two quadratic ones, like x^2+A1x+B1. Then by the assumption, A1^2>B1. Use p/q and q/p,r/s and s/r to establish two equations, and the problem will be down by Cauchy. The inequality holds when p=q=r=s=1. I'm sorry but could you please explain it more clearly i couldn't catch up with you that well :-(

I think my solution is along the lines of what he is doing, except for I used AM-GM to finish.

A post for storage~ (1) When $p=q=r=s$, the equation turns to $x^3+2x^2+2x+1=0$, which means $x=-1$ or $x=\frac{-1\pm \sqrt{3}i}{2}$, so $\lambda \leq \sqrt{3}$. (2) Suppose that when $\lambda = \sqrt{3}$ the result isn’t true. Let $px^3+2qx^2+2sx+r=0$ has roots $a_1+b_1i, a_1-b_1i, r_1$ and the other equation has $a_2+b_2i, a_2-b_2i, r_2$ as roots (2 complex roots and 1 real root), so $|b_1|<\sqrt{3} |a_1|$ and $|b_2|<\sqrt{3} |a_2|$. By Vieta, \begin{align*} p:-2q:2s:-r &= 1:2a_1+r_1:2a_1 r_1+{a_1}^2+{b_1}^2:{a_1}^2 r_1+{b_1}^2 r_1 \\ q:-2p:2r:-s &= 1:2a_2+r_2:2a_2 r_2+{a_2}^2+{b_2}^2:{a_2}^2 r_2+{b_2}^2 r_2, \end{align*}so \begin{align*} \left(\frac{2a_1}{{a_1}^2+{b_1}^2}+\frac{1}{r_1}\right) \left(\frac{2a_2}{{a_2}^2+{b_2}^2}+\frac{1}{r_2}\right) &= 4 \text{ }(1)\\ (2a_1+r_1)(2a_2+r_2) &= 4 \text{ }(2)\\ r_1, r_2 &< 0 \text{ }(3). \end{align*}(just multiply out the ratios) ** When $a_1 \geq 0$, from $0 < 2a_1 r_1+{a_1}^2+{b_1}^2$ and $2a_1+r_1<0$ we get ${b_1}^2 > 3{a_1}^2$. The same goes for $a_2$. ** When $a_1, a_2 < 0$, the first one gives \begin{align*} 4 &> \left( \frac{2a_1}{{a_1}^2+3{a_1}^2}+\frac{1}{r_1} \right) \left( \frac{2a_2}{{a_2}^2+3{a_2}^2}+\frac{1}{r_2} \right) \\ &= \left(\frac{1}{2a_1}+\frac{1}{r_1}\right) \left(\frac{1}{2a_2}+\frac{1}{r_2}\right), \end{align*}so by (2) and Cauchy there’s a contradiction!

Take $p=q=r=s=1$,we got $\lambda \leq \sqrt3$ To prove $\lambda = \sqrt 3$,we suppose the roots of the two equations are $x_1,x_2,x_3;y_1,y_2,y_3$ we have(Viete) $\sum x_i \sum {y_i}=\sum\frac{1}{x_i}\sum\frac{1}{y_i}=4$

$\square$

Thank you,Mr.He.