Question

Let $x$ and $y$ be nonnegative real numbers. If $x^2 + 3y^2 = 18$, then find the maximum value of $x + y$.

Answer 1

[AI-generated answer — Bot GPT 3.5]

We have that $(x + y)^2 = x^2 + 2xy + y^2 = (x^2 + 3y^2) + 2xy = 18 + 2xy.$ By AM-GM,

\[2xy \le x^2 + 3y^2 = 18,\]so $(x + y)^2 \le 18 + 18 = 36,$ which means $x + y \le \sqrt{36} = \boxed{6}.$

Equality occurs when $x = 3y = 6.$