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

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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


`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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