
Re: x + y + z >= xyz implies x^2 + y^2 + z^2 >= Axyz
Posted:
Oct 30, 2013 8:25 AM


quasi wrote: > quasi wrote: >> >> Here's a nice challenge problem which I adapted from a past >> competition problem ... >> >> Problem: >> >> Find, with proof, the largest real number A such that >> >> x + y + z >= xyz >> >> implies >> >> x^2 + y^2 + z^2 >= Axyz >> >> My solution is elementary (avoids Calculus), but if a method >> based on Calculus yields an easy resolution, that would be worth >> seeing as well. > > As a number of people have surmised, the maximum possible > value of A is sqrt(3). > > Here's a proof ... > > Let b = sqrt(3). > > For x = y = z = b, > > x^2 + y^2 + z^2 = b(xyz) > > hence max(A) <= b. > > To show that max(A) = b, it suffices to show that for > all real numbers x,y,z > > x + y + z >= xyz > > implies > > x^2 + y^2 + z^2 >= bxyz > > Suppose otherwise. > > Thus, assume x,y,z are real numbers such that > > x + y + z >= xyz > > x^2 + y^2 + z^2 < bxyz > > First > > 0 <= x^2 + y^2 + z^2 > > => 0 < bxyz > > => xyz > 0 > > => (x + y + z)^2 >= (xyz)^2 > > Next > > (xy)^2 + (yz)^2 + (zx)^2 >= 0 > > => 2*(x^2 + y^2 + z^2) >= 2(xy + yz + zx) > > => 3(x^2 + y^2 + z^2) >= x^2 + y^2 + z^2 + 2(xy + yz + zx) > > => 3(x^2 + y^2 + z^2) >= (x + y + z)^2 > > => 3(x^2 + y^2 + z^2) >= (xyz)^2 > > => 3(bxyz) > (xyz)^2 > > => xyz < 3b > > But > > x^2 + y^2 + z^2 >= 3(xyz)^(2/3) [by AMGM] > > => bxyz > 3(xyz)^(2/3) > > => (bxyz)^3 > 27(xyz)^2 > > => xyz > (27/b^3) > > => xyz > 3b > > contradiction, which completes the proof. > > quasi
That's neat. May I ask where the problem came from, and have you considered similar with more than three numbers?
 What a piece of work is a man. How noble in reason, how infinite in faculties, in form and moving how express and admirable, in action how like an angel, in apprehension how like a God Shakespeare through the mouth of Hamlet

