Question

if a > 0 show that a + 1/a ≥ 2

Answer 1

Answer:

Let a∈R

If a>0, then a+1a≥2

If a<0, then a+1a≤2

This is how someone explained the first one to me but still not really sure about it.

Proof:

⟺a+1a≥2 ⟺ the square of any real number is non-negative so we have (a−1)2≥0 (don't understand this part) ⟺ a2−2a+1≥0 ⟺ a2+1≥2a ⟺ since a>0 then so is a+1a≥2 if a>0

Step-by-step explanation:

Think about it in the other direction: If you square any real number you get a nonnegative result, so

(a−1)2≥0

Expand the left side:

a2−2a+1≥0

If a>0, we divide by a to find

a−2+1a≥0

or upon rearrangement, the desired inequality.

If a<0, division by a reverses the inequality.