quasi
Posts:
10,900
Registered:
7/15/05


Re: x + y + z >= xyz implies x^2 + y^2 + z^2 >= Axyz
Posted:
Oct 30, 2013 4:12 PM


Peter Percival wrote: >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. > >That's neat. May I ask where the problem came from,
I don't remember the source.
It was definitely a competition problem  I think a Junior level contest (but a strong one).
What I remember is that the problem was to prove:
If x + y + z >= xyz then x^2 + y^2 + z^2 >= xyz
From my solution, it became clear that the conclusion could be strengthened from
x^2 + y^2 + z^2 >= xyz
to
x^2 + y^2 + z^2 >= sqrt(3)*(xyz)
>and have you considered similar with more than three numbers?
My solution generalizes directly to prove the following:
For n > 1, the largest real number A such that
x_1 + ... + x_n >= (x_1) * ... * (x_n)
implies
(x_1)^2 + ... + (x_n)^2 >= A*(x_1) * ... * (x_n)
is
A = n^(1/(n1))
quasi

