Let x and y be nonnegative real numbers. If x^2 + 3y^2 = 18, then find the maximum value of x + y.
This might seem like a weird way to do it, but like others, I would reccommend the Cauchy-Schwarz inequality.
Essentially, what this inequality states is that where a and b are real numbers, we can write the inequality (a21+a22+⋯+a2n)(b21+b22+⋯+b2n)≥(a1b1+a1b2+⋯+anbn)2
We could apply this to this problem!
We have (x2+3y2)(1+13)≥(x+√3(1√3)y)2.
18∗43≥(x+y)2
x+y≤√24
Clearly, from this equation, we find that the maximum value is √24
We could probably brute force this problem, but this trick is defintely easier!
Thanks! :)