Maximum Holds When a = b = c = d a=b=c=d , Right?

Algebra Level 3

Let a , b , c a,b,c and d d be real numbers satisfying a 4 + b 4 + c 4 + d 4 = 16 . a^4+b^4+c^4+d^4=16\; . Find the maximum value of a 5 + b 5 + c 5 + d 5 . a^5+b^5+c^5+d^5 \; .


The answer is 32.

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1 solution

Note : This is just a simple version of this problem .

Solution :

a 4 a 4 + b 4 + c 4 + d 4 = 16 a 2 a 5 2 a 4 a^4 \leq a^4 + b^4 + c^4 + d^4 = 16 \Rightarrow a \leq 2 \Rightarrow a^5 \leq 2a^4 . Similarly, it can be proved that b 5 2 b 4 , c 5 2 c 4 b^5 \leq 2b^4, c^5\leq 2c^4 and d 5 2 d 4 d^5 \leq 2d^4 . Adding these up, we get a 5 + b 5 + c 5 + d 5 2 ( a 4 + b 4 + c 4 + d 4 ) = 32 a^5 + b^5 + c^5 + d^5 \leq 2(a^4 + b^4 + c^4 + d^4) = 32 , and equality holds if and only if one of a , b , c , d a,b,c,d is 2 2 and rest all are zero.


[This problem is taken from the book "Inequalities" by Zdravako Cvetovski. It is a nice book, and I recommend it for beginners]

Moderator note:

A more intuitive approach is to use the substitution w = a 4 w = a^4 etc, and see that we want to maximize w 5 4 \sum w^ \frac{5}{4} given w = 16 \sum w = 16 . We can then apply the theory of convex functions to see that the maximum is achieved at the end points.

explain me more clearly, karthik

Arun Garg - 5 years, 2 months ago

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