# Demonstrate the inequality a +1/a>=2

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### 3 Answers

To examine the inequality a+1/a >2.

Solution:

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.

Therefore.

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:

(a-1)^2>=0

But a number square raised it has always a positive value, so the inequality is true!