Find the minimum without Calculus or Graphing

[patrickbotros]

ƒ(ß)=.5sec(ß) + √[1+(sec²(ß)/4)+tan(ß)/√(2)]
Without graphing it or using calculus find the minimum. I already know the answer but want to know how to do it. It s at π/12 and is something like 1.5.
First off this is NOT a homework problem. I already know the answer is something like 1.5 at π/12 but I want to know how to do it. DONT PUT THIS IN A HOMEWORK SECTION.

 

[Dr. Courtney]

Perhaps one can use trig identities to simplify it first.

 

[BvU]

I'd like to know the actual function. Probably not

ƒ(ß)=.5sec(ß) + √[1+(sec²(ß)/4)+tan(ß)/√(2)]

or, as we write it here $$ {1\over 2\cos\beta}+ \sqrt{ 1 + {1\over 4 \cos^2\beta} + {1\over \sqrt 2}\tan\beta} $$ surely ?

 

[nasu]

Probably not.
f(Pi/12)=1.73
f(0)=1.62

@OP
Have you checked if the Pi/12 and 1.5 satisfy the expression?

 

[Dr. Courtney]

Make a large table of values at small step size.

Pick the three lowest adjacent values.

Fit to a quadratic.

Use the standard formula for the minimum of a parabola opening upward.