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Prove that sqrt(1+x)>1+(x/2) using MVT

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should i find f(x) first?
This inequality isn't true, are you sure it's not sqrt(1+x)<1+(x/2)? For example if you try x=1 sqrt(1+1)>1+1/2 => sqrt(2)>1.5 is false.
yes i wrote it wrong, the sign is the wrong way

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im just not sure how to apply the MVT
square both sides and see what you get .
You can square both sides but it says mean value theorem? Does the question give you some lower bound like for all x>0 or x>=-1? As the mean value theorem says: for a differentiable function f(x) on the closed interval (a,b) and is continuous on [a,b] then there exists a c where a<=c<=b where: \[f \prime(c)=\frac{ f(b)-f(a) }{ b-a }\] and you are correct that in this case f(x)=1+(x/2)-sqrt(1+x).
ya i cant square both sides
ty quantum, didn't even know what MVT were :), sorry calle87 :)
not sure how to do it with just {a,b} use to having an interval
s'ok :}
f'(c) is strange
Okay so that means that there exists a c where 00: 1/sqrt(1+c)<1 therefore should be able to work out f'(c)>0 for all c>0. Then, \[\frac{ 1+x/2-\sqrt(1+x) }\] Since, x>0 the numerator is also greater than 0. Which is the original inequality :)
Sorry in the last equation it should have said:\[\frac{ 1+x/2-\sqrt(1+x) }{ x }>0\]
hmmm, f'(c) why is the one not zero?
this is were i get lost
Sorry (shoun't be a one). Let me work it out :p
Haha, we both messed up the derivative... it should be f(c)=1/2*(1-1/sqrt(1+c)). Which is not hard to show it is greater than 0
\[\sqrt(1+c)>1 => 1/\sqrt(1+c)<1 => 0.5*(1-1/\sqrt(1+c))>=0\]
so c is a constant
c is some constant in the range (0,x)
mahmit how do you get that?
what are you doing? it is solved but with different inequality!!
so i can just take sqrt(1+x) as f(x) and apply the MVT to it directly
wow omg, thank you so much!
but instead of > should be<
unless for x<0
well what do i do with this?
wait how do you just jump from here ?
@mahmit2012 do you know what happens to this?

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