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Prove that $x-\frac{1}{x^2} \geq 0$ for every $x \ge 1$

I know that this can be done using elementary algebra, but in other cases it's not that simple. Can I prove this inequation positive for every $x \ge 1$ if I show that it's positive for the smallest value, that's 1 in this case and then that limit of the function $f(x) = x-\frac{1}{x^2}$ as x approaches infinitiy is infinity?

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Er... do you mean $\geq 0$? Because when $x = 1$... –  Sp3000 Jun 17 '13 at 15:46
    
Yes it should be $\geq 0$. My bad –  Stefan4024 Jun 17 '13 at 17:33

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No, you can't. There are functions that start and end positive but are negative somewhere in the middle. On the other hand, if you could show that your function is increasing...

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No, unfortunately you can't. Consider the function $y=x(x-1)(x-2)$ as a counterexample.

However, try looking at the derivative of the function as this will tell you how the function changes over time and hence if it will stay non-negative.

P.S. You may want to check your inequalities as currently your statement isn't true for $x=1$

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I don't understand why your function is counterexample, for every $x>2$ it's bigger than 0. –  Stefan4024 Jun 17 '13 at 17:35
    
But not for x between 1 and 2 or less than 0. –  marty cohen Jun 17 '13 at 18:02
    
the function was chosen to closely match your example. It is 0 when $x=1$ and tends to infinity as x tends to infinity but it is not greater than or equal to zero for all $x\ge1$ as it is negative for $1<x<2$ –  john Jun 18 '13 at 2:41

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