a*b*c*d*e = 1 & a^4 + b^4 + c^4 + d^4 + e^4 ≥ a + b + c + d + e

 
 
Given   a, b, c, d, e > 0,           a\cdot b \cdot c\cdot d \cdot e = 1,

show that

a^4 + b^4 + c^4 + d^4 + e^4 \; \geq \; a + b + c + d + e

Find as many diferent solutions as you can.
 
 

 
 
 
 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.

2 Responses to a*b*c*d*e = 1 & a^4 + b^4 + c^4 + d^4 + e^4 ≥ a + b + c + d + e

  1. paul says:

    There will be an infinite numbers of solutions to this, if we set a, b, c, d to the reciprocals of the numbers 2, 3, 4, 5 then to make a b c d e = 1 we need e to be the product of those numbers i.e. 120, raising those numbers to the fourth power and summing them will always be greater than the sum of them without the power, the reciprocals of those given numbers makes the total of fourth powers the smallest it can be. As there are an infinite set of numbers there will be an infinite solution set.

    Paul.

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