How do I prove that x + 1/x is greater than or equal to 2 if x > 0

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To prove that x + 1/x is greater than or equal to 2 for x > 0 without using calculus, one can start by rearranging the expression into the form x^2 - 2x + 1 ≥ 0. This simplifies to (x - 1)^2 ≥ 0, which is always true for any real number x. Since x is constrained to be greater than 0, the inequality holds, confirming that x + 1/x is indeed greater than or equal to 2. Additionally, using the arithmetic-geometric mean inequality reinforces this conclusion, as it leads to the same result. Thus, the proof is established through algebraic manipulation and properties of inequalities.
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How to prove this?

How do I prove that x + 1/x is greater than or equal to 2 if x > 0

i'm not allowed to use calculus either.

i got that x + 1/x is greater than zero, but i can't get greater than or equal to 2.
 
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Assume for contradiction that x > 0 yet x + 1/x < 2. Transform it into a quadratic and show that this is impossible.
 
Hint: (x - 1)^2 > 0.

Regards,
George
 
First make the equation into x^2-2x+1 >= 0 by rearranging and multiplying by x.

You should find it is (x-1)^2 so for any x (not just x>0) this function is greater than or equal to 0 because of ^2.

For your situation it is obvious that x>0 which is certainly true for the factorised function. Therefore you have proved that x+1/x is greater than or equal to 2.
 
pivoxa15 said:
First make the equation into x^2-2x+1 >= 0 by rearranging and multiplying by x.

You should find it is (x-1)^2 so for any x (not just x>0) this function is greater than or equal to 0 because of ^2.

For your situation it is obvious that x>0 which is certainly true for the factorised function. Therefore you have proved that x+1/x is greater than or equal to 2.

before i posted this, i did get the algebraic manipulation of (x-1)^2, but i thought that only proved it for x greater than or equal to 1. but i guess since it's an equivalent statement, it's the same thing.

thanks everyone
 
Another method is arithmetis-geometric mean inequality.
 
Strictly speaking a proof would work the other way:

for any x, (x-1)^2\ge 0 so x^2- 2x+ 1\ge 0.
Adding 2x to both sides, x^2+ 1\ge 2x. Finally, dividing both sides by the positive number x, x+ \frac{1}{x}\ge 2
 
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