Question

if a and b are two positive real number then their A.M. is grather than or equal to their G.M.

Answer 1

proof, we know that (√a-√b)²≥0=>a+b-2√ab≥0
=> a+b≥2√ab=>a+b/2≥√ab proved.
Note. if a≠b then a+b/2 > √ab ; if a = b then a+b/2=√ab

Answer 2

if two positive numbers a and b are given then
Arithmetic mean of a and b is always greater then geometric mean of a and b
e.g AM ≥ GM
Here,
a ≥ 0 and b ≥ 0
Also we can say that
(√a - √b)² ≥ 0 for all real number of a and b
$\sqrt{a}^2 + \sqrt{b}^2 -2\sqrt{a}\sqrt{b}$≥0
a + b ≥ 2√ab
(a + b)/2 ≥√ab
Hence AM ≥ GM