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

Pythagorearn May 27, 2024

#1**+1 **

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! :)

NotThatSmart May 27, 2024