Question

Prove that sec^2x+cos^2x can never be less than 2

Answer 1

Since, Arithmatic mean ≥ Geometric Mean
Then for any x,y we have,
(x+y) ≥ 2√xy
Let, x=a and y=1/a
∴, (a+1/a) ≥ 2√(a×1/a)
or, a+1/a ≥ 2
Putting, a=cos²x we get,
cos²x+1/cos²x ≥ 2
or, cos²x+sec²x ≥ 2 (Proved)

Answer 2

Derivatives method:

let   cos x = y ,      sec x = 1/y
Let  M = y^2 + 1/y^2
dM/dy = 2 y - 2/y^3    = 0    =>   y = +1  or -1

d^2 M/dy^2 = 2 + 6 / y^4   > 0
        so  at  y = 1 or -1, (at x=0) , we have a minimum value for M.

Minimum value of M = 1 + 1 = 2