Basic mathematics problem # 1
If x,y,x be positive real numbers such that x 2 + y 2 + z 2 = 2 7 , then what is the minimum value of x 3 + y 3 + z 3 ?
The answer is 81.
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2 solutions
Just to clarify, (5,1,1) would also satisfy the first equation but not the second so you are correct
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By AM-GM,
x 2 + y 2 + z 2 ≥ 3 3 x 2 y 2 z 2
3 x 2 y 2 z 2 ≤ 2 7
Equality holds iff x = y = z = 3
3 x 2 y 2 z 2 ≥ 9
x 3 + y 3 + z 3 ≥ 3 x y z ≥ 8 1
Checking, we can see that 8 1 is indeed the maximum.
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Not sure how you have 3 3 x 2 y 2 z 2 = 2 7 .
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@Mathh Mathh – Oops, typo. Thanks! (If "Why 27?" was the question, it's because it was given.)
Thanks, BTW Why don't you post it as a solution?
@Sean Ty 2 7 ≥ 3 3 x 2 y 2 z 2 !!!
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@Samuraiwarm Tsunayoshi – Okay, okay, I'll change this now. I have a seminar now. Peace.
Very nice, but can directly be solved by Power Mean inequality.
shouldnt it be -81
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Read the Q carefully!! It says x,y and z are positive. Thus the sum of their cubes will also be positive. Hence 81.
x³+y³+z³=(x+y+z)(x²+y²+z²-xy-yz-zx) + 3xyz now x+y+z and 3xyz are always greater than 0 as given that x,y,z>0 Now minimum will be when x²+y²+z²-xy-yz-zx=0 solving x=y=z Now x²+y²+z²=27 put x=y=z We get x=y=z=3 this is the condition for minimum value. Solve x³+y³+z³=81
x=y=z=3 is the only set of nos. satisfying the 1st eqn. Hence cube of each x,y,z results in 27*3=81