Maximum
a 5 + b 5 + c 5 + d 5
If a , b , c , d are real numbers such that a 4 + b 4 + c 4 + d 4 = 1 6 , find the maximum value of the expression above.
The answer is 32.
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1 solution
What's wrong in this?
4 ( a 5 ) 5 4 ( b 5 ) 5 4 ( c 5 ) 5 4 ( d 5 ) 5 4 ≥ ( 4 a 5 + b 5 + c 5 + d 5 ) 5 4
Solving this we get Max as 2 2 9
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I am also thinking on the same path
Can you get values of a , b , c , d such that the LHS = RHS? See for a 5 + b 5 + c 5 + d 5 = ( a 4 + b 4 + c 4 + d 4 ) 4 5 , the vaues are ( 0 , 0 , 0 , 2 ) .
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But sir, (0,0,0,2) is satisfying the LHS and RHS a 5 + b 5 + c 5 + d 5 = ( a 4 + b 4 + c 4 + d 4 ) 4 5
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@Md Zuhair – My solution is valid because, there is ( a , b , c , d ) satisfying a 5 + b 5 + c 5 + d 5 = 3 2 . You cannot find ( a , b , c , d ) satisfying a 5 + b 5 + c 5 + d 5 = 2 2 9 , therefore, the solution is not valid.
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@Chew-Seong Cheong – O, Now i got it sir. Thanks :)
Can the value of 32 be obtained?
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Thanks, I have added in the solution.
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Wrong. When a = b = c = d , then the value of a 5 + b 5 + c 5 + d 5 is 1 6 2 , not 32.
The value of 32 is obtained when (a,b,c,d) is a permutation of (2,0,0,0)?
Why the sum of exponent on the right hand side does not equal to 1? Does holder inequalities still work?
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You are right. I need to recheck this. Thanks.
Using Hölder's inequality as follows:
a 5 + b 5 + c 5 + d 5 ≤ ( a 4 + b 4 + c 4 + d 4 ) 4 1 ( a 4 + b 4 + c 4 + d 4 ) 4 1 ( a 4 + b 4 + c 4 + d 4 ) 4 1 ( a 4 + b 4 + c 4 + d 4 ) 4 1 ( a 4 + b 4 + c 4 + d 4 ) 4 1 = ( a 4 + b 4 + c 4 + d 4 ) 4 5 = 1 6 4 5 = 3 2
Equality occurs when ( a , b , c , d ) = ( 0 , 0 , 0 , 2 ) .