Let x and y be nonnegative real numbers. If x^2 + 3y^2 = 18, then find the maximum value of x + y.
There's a really good inequality for this, Cauchy-Schwarz inequality.
(a21+a22+⋯+a2n)(b21+b22+⋯+b2n)≥(a1b1+a1b2+⋯+anbn)2 , where a and b are a sequence of real numbers.
Applying this, we want to get xy on the right side so we do:
(x2+3y2)(1+13)=(x+√3(1√3)y)2
18∗43≥(x+y)2
x+y≤√24.
The maximum value is √24.
(Yes the equality condition can be satisfied).