## Global optimization subject to constraints

1a) Determine the maximum value of f(x,y,z)=(xyz)1/3 given that x,y,z are nonnegative numbers and x+y+z=k, k a constant.

1b) Use the result in (a) to show that if x,y,z are nonnegative numbers, then (xyz)1/3 < (x+y+z)/3

Attempt:
1a) Using the Lagrange Multiplier method, I get that the absolute maximum of f subject to the constraints x+y+z=k and x,y,z>0 is k/3

1b) Here, it seems to me that one of the constraints, namely x+y+z=k, is removed. If so, then how can we still use the result of part (a) here?

I need some help on part (b). Any help is appreciated!
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 Quote by kingwinner 1a) Determine the maximum value of f(x,y,z)=(xyz)1/3 given that x,y,z are nonnegative numbers and x+y+z=k, k a constant. 1b) Use the result in (a) to show that if x,y,z are nonnegative numbers, then (xyz)1/3 < (x+y+z)/3 Attempt: 1a) Using the Lagrange Multiplier method, I get that the absolute maximum of f subject to the constraints x+y+z=k and x,y,z>0 is k/3 1b) Here, it seems to me that one of the constraints, namely x+y+z=k, is removed. If so, then how can we still use the result of part (a) here? I need some help on part (b). Any help is appreciated!
For any x, y, z, x+ y+ z is something isn't it? For any x, y, z, define k= x+ y+ z. Then you have shown by a that $(xyz)^{1/3}\le k/3= (x+ y+ z)/3$.

 Quote by HallsofIvy For any x, y, z, x+ y+ z is something isn't it? For any x, y, z, define k= x+ y+ z. Then you have shown by a that $(xyz)^{1/3}\le k/3= (x+ y+ z)/3$.
For simplicity, let's take k=5.

In part b, x,y,z are only required to be nonnegative numbers. There is no restriction that x+y+z=5 as there is in part a.
Take e.g. x=5, y=5, z=5 which are nonnegative
But x+y+z=15, which is not equal to 5.

It seems to me that (xyz)1/3 < (x+y+z)/3 is true only if x+y+z=k, but NOT true for ANY nonnegative numbers, and in part b we have to prove the latter.

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