How √x + 1/√x >2 or = 2 for x>0
we have since all squares are positive
(a-b )^2 >= 0
expanding we have
a^2+b^2 > 2ab
from the question we can see that we are only considering positive values of √x
so we can safely
assume √x=t^2
1/√x=1/t^2
√x + 1/√x= t^2+1/t^2
now let a= t , b= 1/t
we have t^2+1/t^2 > 2t*1/t
or
t^2+1/t^2 >= 2
gives
√x+1/√x >=2
Since x>0 and √x >0.
√x + 1/√x >= 2
1/√x >= 2 -√x
1 >= 2√x -x
-1 <= x-2√x = (√x)^2-2√x
0 <= (√x)^2 -2√x -1 = ( √x -1)^2
And (a)^2 >= 0 is always true ( for a real Number)
So for x>0 the following is true:
√x + 1/√x >= 2 is equivalent to 0 <= ( √x -1)^2, which is always true (for x real)
qed
Let x > 0 (so 1/√x is well defined and real). Note that as a square
0 ≤ (x^(1/4) - x^(-1/4))² = √x - 2 + 1/√x ==> √x + 1/√x ≥ 2.
*** P.S. The square root is just to make the problem appear more complicated then it actually is. It is in fact true that for all t > 0,
t + 1/t ≥ 2.
It is not necessary that t ≥ 1, only that t is positive. Equality holds only in the case t = 1.
Let √x = t then t >0 and t + 1/t - 2 =(1/t)(t^2 +1 - 2t) = (1/t) (t - 1 )^2 > or = 0 for all values of t >0.
so √x + 1/√x >2 or = 2 for x > 0.
multiply everything by √x
x + 1 = 2√x
re-arrange to make = 0
x - 2√x + 1 = 0 (this is a quadratic in the form u^2 - 2u + 1 where u = √x)
Factorise
(√x - 1)(√x - 1) = 0
x = 1 (one real root)
√x + 1/√x >2 or = 2
simplify
x+1>2 √xor = 2√x
x-2√x+1>0 or= 0
(√x-1)^2>0 or =0
(√x-1)>0 or= 0
√x>1or= 1
square brackets -1,1
get free math homework help at http://www.mathskey.com/
Because the lowest number you can put as x is 1
So if you use 1 :
sqrt1 + 1/squrt1 = 2 (See? The lowest answer you can get is 2 so that's why it can be equal to 2)
Then if you put other numbers like 4
it will be sqrt4 + 1/sqrt4 = 2+1/2 = 2.5 (Now the larger number you put, the larger number you get as the answer so that's why it says as >2
So √x + 1/√x >= 2
Hope it help :)
√x + 1/√x ≥ 2......multiply across by √x.....
x + 1 ≥ 2√x
x - 2√x ≥ -1............This is true for all positive values of x
Copyright © 2024 1QUIZZ.COM - All rights reserved.
Answers & Comments
Verified answer
we have since all squares are positive
(a-b )^2 >= 0
expanding we have
a^2+b^2 > 2ab
from the question we can see that we are only considering positive values of √x
so we can safely
assume √x=t^2
1/√x=1/t^2
√x + 1/√x= t^2+1/t^2
now let a= t , b= 1/t
we have t^2+1/t^2 > 2t*1/t
or
t^2+1/t^2 >= 2
gives
√x+1/√x >=2
Since x>0 and √x >0.
√x + 1/√x >= 2
1/√x >= 2 -√x
1 >= 2√x -x
-1 <= x-2√x = (√x)^2-2√x
0 <= (√x)^2 -2√x -1 = ( √x -1)^2
And (a)^2 >= 0 is always true ( for a real Number)
So for x>0 the following is true:
√x + 1/√x >= 2 is equivalent to 0 <= ( √x -1)^2, which is always true (for x real)
qed
Let x > 0 (so 1/√x is well defined and real). Note that as a square
0 ≤ (x^(1/4) - x^(-1/4))² = √x - 2 + 1/√x ==> √x + 1/√x ≥ 2.
*** P.S. The square root is just to make the problem appear more complicated then it actually is. It is in fact true that for all t > 0,
t + 1/t ≥ 2.
It is not necessary that t ≥ 1, only that t is positive. Equality holds only in the case t = 1.
Let √x = t then t >0 and t + 1/t - 2 =(1/t)(t^2 +1 - 2t) = (1/t) (t - 1 )^2 > or = 0 for all values of t >0.
so √x + 1/√x >2 or = 2 for x > 0.
multiply everything by √x
x + 1 = 2√x
re-arrange to make = 0
x - 2√x + 1 = 0 (this is a quadratic in the form u^2 - 2u + 1 where u = √x)
Factorise
(√x - 1)(√x - 1) = 0
x = 1 (one real root)
√x + 1/√x >2 or = 2
simplify
x+1>2 √xor = 2√x
x-2√x+1>0 or= 0
(√x-1)^2>0 or =0
(√x-1)>0 or= 0
√x>1or= 1
square brackets -1,1
get free math homework help at http://www.mathskey.com/
Because the lowest number you can put as x is 1
So if you use 1 :
sqrt1 + 1/squrt1 = 2 (See? The lowest answer you can get is 2 so that's why it can be equal to 2)
Then if you put other numbers like 4
it will be sqrt4 + 1/sqrt4 = 2+1/2 = 2.5 (Now the larger number you put, the larger number you get as the answer so that's why it says as >2
So √x + 1/√x >= 2
Hope it help :)
√x + 1/√x ≥ 2......multiply across by √x.....
x + 1 ≥ 2√x
x - 2√x ≥ -1............This is true for all positive values of x