Question

If p q r and s are real numbers such that pr equal to 2 + us then show that at least one of the equation

Answer 1

There is a little correction in the above instead of rrthe author wrote qq. Given: pr=2(q+s)pr=2(q+s). To prove: either p2−4q≥0p2−4q≥0 or r2−4s≥0r2−4s≥0. It is best to argue by contradiction -- assume both p2−4q<0p2−4q<0 and r2−4s<0r2−4s<0. Then upon rearranging and adding the inequalities, p2+r2<4q+4sp2+r2<4q+4s p2+r22<2(q+s)p2+r22<2(q+s) p2+r22<prp2+r22<pr (pr=2(q+s))(pr=2(q+s)) (p2+r2)<2pr(p2+r2)<2pr p2+r2−2pr>0p2+r2−2pr>0 (p−r)2>0(p−r)2>0 it means p=rp=r. –