Logs And inequalities

Question

x, y and n are positives

Whats the minimum value of xⁿ + yⁿ - (xy)^(n/4)?

Answer 1

Replacing x^n/4 and y^n/4 with a and b, respectively, the expression becomes a⁴ + b⁴ - ab. Minimizing the given expression is the same as minimizing a⁴ + b⁴ - ab.

The partial derivative with respect to a is 4a³ - b. The partial derivative with respect to b is 4b³ - a. Setting these equal to zero and solving the resulting system gives three critical points (-1/2,-1/2), (0,0), and (1/2,1/2). The first two critical points are rejected since a and b must be positive. The last critical point gives the minimum value of -1/8. You can be sure this is the minimum because the second partial derivatives, 12a² and 12b², are always positive.

Answer 2

We like to think that N is an integer and usually it is, but since it NOT explicitly given,

we cannot assume it is.

This is a NASTY calculus problem.....

F(x,y) = x^n + y^n - x^(n/4)*y^(n/4)

0 = dF/dx = n*x^(n-1) - (n/4)*x^(-3n/4) * y^(n/4)

0 = dF/dy = n*y^(n-1) - (n/4)*y^(-3n/4)*x^(n/4)

solves the first equation for y:

n*x^(n-1)/(n/4)*x^(-3n/4) = y^(n/4)

4 * x^( 3n/4 - 1) = y^(n/4)

Takes log of both sides

ln 4 + (3n/4 - 1)* ln x = (n/4) ln y

(4/n) ln 4 + (4/n) ( 3n/4 -1) * ln x = ln y

exp ( (4/n) ln 4 + (3-4/n)*ln x) = y

n*exp ( (4/n) ln 4 + (3-4/n)*ln x)^(n-1) - (n/4)*exp ( (4/n) ln 4 + (3-4/n)*ln x)^(-3n/4)*x^(n/4) = 0

you now need to graph this horrible function to find the irrational solutions

please repost