# How to solve 3^sin^6x + 3^cos^6x by A.m. G.m. Method?

Hello!

The AM-GM inequality states that for any non-negative numbers a and b

(a+b)/2gt=sqrt(a*b), or a+bgt=2sqrt(a*b).

Therefore it may be suitable for estimating a sum from the below.

Denote sin^2(x)=u, then cos^2(x)=1-u and the function becomes

3^(u^3)+3^((1-u)^3).

u may be any number in [0,1].

We may apply this for our function and obtain an inequality

3^(u^3)+3^((1-u)^3) gt= 2sqrt(3^(u^3)*3^((1-u)^3)) =

= 2*3^(1/2(u^3+(1-u)^3)) = 2*3^(1/2(1-3u+3u^2)).

The exponent 1/2(1-3u+3u^2)  has one and only one minimum at u_0=1/2 in [0.1], the value at u_0 is 1/8.

So for any u we have 3^(u^3)+3^((1-u)^3) gt= 2*3^(1/8), and this inequality becomes an equality only at u_0=1/2.

Now recall that  u=sin^2(x). It is equal to 1/2 when sin(x)=+-1/sqrt(2), so at x=pi/4+(k pi)/2. This way we have found minimums of the given function. They are pi/4+(k pi)/2, and the minimum value is 2*3^(1/8) approx 2.294.

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