Let x, y, and z be real numbers such that x + y + z = 1, and x is greater than or equal to -1/3, y is greater than or equal to -1 and z is greater than or equal to -5/3. Find the maximum value of \(\sqrt{3x+1}+\sqrt{3y+3}+\sqrt{3z+5}\)
+321 I will attempt this problem, but I'm not sure whether or not my answer is right or not. I set x, y, and z all equal and I got sqrt(2) + 2 + sqrt(6).
