Proving inequality by induction,given a condition

realanony87
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Homework Statement


If x_{1} x_{2} \cdots x_{n}=1 (1)
show that
x_{1}+x_{2}+\cdots+x_{n} \geq n (2)

The Attempt at a Solution



I attempted as follows. I started with

x_{1} + \frac{1}{x_{1}} \geq 2 , which is an inequality I already know how to prove.

Then using Eq.(1) I get
x_{1} + x_{2} x_{3} \cdots x_{n} \geq 2

Continuing from this point , for example started from another point x_{2} and repeating the procedure for all n , I get no where. I cannot think of another path to take.

If i try to do it by induction, I cannot assume that the equation holds for n numbers , and try to prove for n+1 numbers, as by including x_{n+1}, Eq.(1) and Eq.(2) need not hold anymore but
x_{1} x_{2} \cdots x_{n} x_{n+1}=1
x_{1}+x_{2}+\cdots+x_{n} +x_{n+1}\geq n +1

Edit:
Assuming all x's are nonnegative
 
Last edited:
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I'm assuming all the x_i are nonnegative real numbers.

Do you really want to do this by induction? It follows pretty easily from the AM-GM inequality.
 
Alternatively, you can use a Lagrange multiplier to extremize the sum of the xi's subject to the constraint that their product is 1.
 
Well I could use AM-GM , or the method by langrange multiplier but I am interested in how to apply the principle of induction itself when such a condition is given, or if its possible at all .
 
I don't really see how to get at this with induction.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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