## 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.

math grad - Interest: Number theory
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### 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.