+415 Let x and y be nonnegative real numbers. If x^2 + 3y^2 = 18, then find the maximum value of x + y.
+1365 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 \((a_1^2+a_2^2+\dots+a_n^2)(b_1^2+b_2^2+\dots+b_n^2)\ge(a_1b_1+a_1b_2+\dots+a_nb_n)^2\)
We could apply this to this problem!
We have \((x^2+3y^2)(1+\frac{1}{3}) \geq (x+\sqrt{3}(\frac{1}{\sqrt{3}})y)^2\).
\(18*\frac{4}{3}\ge(x+y)^2\)
\(x+y\le\sqrt{24}\)
Clearly, from this equation, we find that the maximum value is \(\sqrt{24}\)
We could probably brute force this problem, but this trick is defintely easier!
Thanks! :)
