Proving x+1/x >= 2 for x>0 in Math: Helpful Tips and Tricks

  • Thread starter Thread starter mercury
  • Start date Start date
AI Thread Summary
To prove that x + 1/x ≥ 2 for all x > 0, one can start by manipulating the inequality into the form x² - 2x + 1 ≥ 0, which is a quadratic that is always non-negative. This can be confirmed by recognizing that the expression factors into (x - 1)², indicating equality at x = 1 and positivity elsewhere. Additionally, the Arithmetic Mean-Geometric Mean (AM-GM) inequality can be applied, reinforcing that the arithmetic mean of two positive numbers is always greater than or equal to their geometric mean. The discussion emphasizes the importance of starting from known truths and working backwards to derive the desired inequality. Overall, the inequality holds true for all x > 0.
mercury
ok i have a really stupid problem ...but i can't seem to get it...would appreciate any help...
how do u prove that x+1/x >= 2 for all x>0...
is the above state ment true in the first place..i think it is...but can't seem to prove it...i keep coming to the fact that it holds for x>1 and equality holds for x=1 but can't seem to prove it...obviously for 0<x<1...
 
Physics news on Phys.org
Since you are specifying that x> 0, you can multiply through the inequality by x without any problem:

You want to get x+ 1/x>= 2 and, multiplying by x, that is the same as x2+ 1>= 2x.

That, of course, is the same as x2- 2x+ 1>= 0 and you KNOW that's true.

Well, HOW do you know it's true? Also remember you have to ARRIVE at "x+ 1/x>= 2", not start there. Start with what you KNOW is true and work backwards.
 
Another method is to use the fact that A.M. > = G.M. as x > 0

Where AM stands for arithmatic mean and GM stands for Geometric mean.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top