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To examine the inequality a+1/a >2.
Let f(a) =a+1/a >=2. This implies,
(a^2-2a+1)^2/a >=0 or
f(a) = (a-1)^2/a >=0
Now the numerator is positive for all a ,postive or negative and zero when a= 1. But the denominator is positive only for a>0. Therefore, f(a) is positive only for a>0, and equals to zero when a = 1.
Or f(a) = a+1/a >= 2 for all x>0 and
At a = 0+ f(0+) = +inf
At a=0- , f(0-) =-inf.
So at a = 0, f(a) undefined and excluded.
a+1/a >=2 is only valid for a > 0 and does not hold for x < or = 0. or
a+1/a remains undefined for x=0.So,a+1/a >2 does not hold for x = 0.
a+1/a < 2 for x<0. So a+1/a >2 doesnot hold when x<0
The expression a + 1/a >= 2 means that minimum value of the function f(a) = a + 1/a is 2.
The minimum value of function f(a) occurs for value of a for which derivative f'(a) = 0
So we work out derivative f'(a) and equate it to 0 as follows.
f('a) = 1 - 1/(a^2) = 0
Therefore: 1/(a^2) = 1
Therefore: a^2 = 1
Therefore: a = 1
Putting this value of a in the function f(a) we get minimum value of f(a) as
f(1) = 1 + 1/1 = 2
Therefore given equality is correct.
To prove the inequality [a +(1/a)]>=2, first we have to amplify the integer terms with the only denominator in the inequality, which is "a":
a*a + 1 >=2*a
We'll move the term from the right side of the inequality, to the left side, in order to obtain an expression E(a)>=0 and to verify its true value.
a^2 +1 -2*a>= 0
We'll order decreasingly, considering the powers of "a".
a^2-2*a +1 >= 0
Wenotice that the obtained expression could be written as:
But a number square raised it has always a positive value, so the inequality is true!
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