Expert: Sherry Wallin - 3/29/2009

Question
QUESTION: Hello! I've the following problem: x, y are positive numbers and x+y=1. How can i proof, this inequality  x*y^(2/3)<7/3 ?

thanks

ANSWER: Grull~
   There are a number of ways to prove this. One way is to use the fact that if both x and y are positive and their sum is 1 then both x and y are < 1, i.e., 0<x<1 and 0<y<1. Any number between 0 and 1 taken to a power is also less than 1 and any product of two numbers such as this x and y is < 1. Symbolically we have: x*y^(2/3)< 1*1=1<7/3

If you need to show that x and y are less than 1 here is a way to show it:

using x+y = 1 then x = 1-y but x > 0 so 1-y > 0 --> 1 > y  -->  y < 1 similarly you can show that x < 1.

Math Prof

---------- FOLLOW-UP ----------

QUESTION: Hello, terrible sorry, but i've misstyped the right side, so the correct version is: x*y^(2/3)<3/7.

Answer
I'm hoping you are in at least calculus. Take the derivative of
f(x) = x(1-x)^(2/3):
x(2/3)(1-x)^(-1/3)(-1)+(1-x)^(2/3)(1) = (-2/3)x(1-x)^(-1/3)+ (1-x)^(2/3)
factor out (1-x)^(-1/3)getting:
(1-x)^(-1/3)(-(2/3)x+ 1-x) = (1-x)^(-1/3)(1-(5/3)x) = 0 so 1-x = 0 ->
x = 1 but x is always smaller than 1 and 1-(5/3)x = 0 -> 1 = (5/3)x -> x = 3/5. Now plug x = 3/5 into f(x) and get (.6)(1-.6)^(2/3) = .6(.4)^(2/3) approx .326 and 3/7 is approx .429 so f(x) is <= to 3/7   

Math Prof

Notice at x = 0 or 1 which this function can never take on because of the x+y =1 and both x and y positive, even if that were not true f(0)=f(1) = 0 which is certainly less than 3/7 also. What we used here is the local maxima on the interval [0,1] but since we exclude 0 and 1 it is really the open interval (0,1)