If c> 0, prove that c+1/c is greater than or equal to 2

by algebra, we combine the fractions: c + 1/c = c²/c + 1/c = (c²+1)/c ≥ 2.

(c²+1)/c ≥ 2 and we multiply both sides by c. c is positive so inequality sign stay s.

c² + 1 ≥ 2c ⇔ c²-2c + 1 = (c - 1)(c - 1) ≥ 0

consider case where 0 < c < 1. by the subtraction property of equality, subtracting through by 1, we have

-1 < c - 1 < 0

so c - 1 is negative for 0 < c < 1. but (c - 1)(c - 1)   means that we have a negative times a negative which is a positive. thus (c - 1)(c - 1) ≥ 0 is true for 0 < c < 1.

consider case where c = 1. then (1-1)(1-1) = 0 ≥ 0 which is true

consider the case where c > 1. by the subtraction property of equality, this is equivalent to c - 1 > 0. so c - 1 is a positive number, and (c-1)(c-1) is positive, thereby satisfying (c - 1)(c - 1) ≥ 0.

so c + 1/c ≥ 2 is true for all c > 0.

answer: haha (: i’ll go see what it is

step-by-step explanation: