Suppose each of x and y is a real number > 0
Find the minimum value of:
2x2+2y2+3xy+1
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x+y
I'll start with the obvious, take both first partial derivatives:
d/dx = (2x^2+4xy+y^2-1)/(x+y)^2
d/dy = (x^2+4xy+2y^2-1)/(x+y)^2
A critical point will occur when both partial derivatives equal 0. Since the domain is positive reals, this means I can focus on only the numerator as the denominator is strictly positive over the domain.
Then 2x^2+4xy+y^2-1=0 and x^2+4xy+2y^2-1=0.
Taking the difference of these equations reduces down to x^2-y^2=0, which factors down to x-y=0 or x+y=0.
The x+y=0 branch can be discarded as it is outside the domain. That leaves x-y=0, or x=y. Then back-substituting gets us 7x^2-1=0, which has one positive root x=1/sqrt(7).
Then (x.y)=(1/sqrt(7),1/sqrt(7)) is a critical point. The original expression evaluated at (1/sqrt(7),1/sqrt(7)) is (2/7+2/7+3/7+1)/(2/sqrt(7)) = sqrt(7).
Exploration of the neighborhood, similar to Larry did, shows this to be a minimum.
Solution
