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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 \(\sqrt{3x+1}+\sqrt{3y+3}+\sqrt{3z+5}\)

 Oct 28, 2019
 #1
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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).

 Oct 28, 2019
 #2
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+1

x = 1

y = 1

z =-1

2 + sqrt(6) + sqrt(2) =~5.864

The same answer as "TOP"

 Oct 28, 2019

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