+1839 Let x and y be nonnegative real numbers. If x^2 + 3y^2 = 18, then find the maximum value of x + y.
+399 There's a really good inequality for this, Cauchy-Schwarz inequality.
\((a_1^2+a_2^2+\dots+a_n^2)(b_1^2+b_2^2+\dots+b_n^2)\ge(a_1b_1+a_1b_2+\dots+a_nb_n)^2\) , where a and b are a sequence of real numbers.
Applying this, we want to get xy on the right side so we do:
\((x^2+3y^2)(1+\frac{1}{3})=(x+\sqrt{3}(\frac{1}{\sqrt{3}})y)^2\)
\(18*\frac{4}{3}\ge(x+y)^2\)
\(x+y\le\sqrt{24}\).
The maximum value is \(\sqrt{24}\).
(Yes the equality condition can be satisfied).
