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prove that if x ≥ 0, then 1+x ≥ 2√x

 
 Oct 10, 2021
 #1
avatar+2246 
+2

Since x>= 0, we can square this inequality on both sides. 

 

1 + x >= 2sqrt(x)

x^2 + 2x + 1 >= 4x

x^2 - 2x + 1 >= 0

(x-1)^2 >= 0

A square number always has to be non-negative. 

 

=^._.^=

 
 Oct 10, 2021
 #2
avatar+769 
+1

spitting facts

 

i think the problem should state that this holds for all real x

 
CentsLord  Oct 10, 2021
 #3
avatar+2246 
+1

I'm a bit curious, does the x >= 0 imply that x is a real number?

Since something like 3i can't be greater than 0 right?

 

=^._.^=

 
catmg  Oct 10, 2021
 #4
avatar+769 
+1

yeah, inequalities with complex numbers aren't well defined lol

i guess the x >= 0 would imply that

 
CentsLord  Oct 10, 2021
 #5
avatar+2246 
+1

I don't know if it's even possible to say that one complex number is greater than another. 

I guess you could say like |3 + 4i| = 5 and |12i + 5| = 13, and 5 < 13. (basically like further away from origin)

 

=^._.^=

 
catmg  Oct 10, 2021

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