How demonstrate cos x < cos(x/3)cos(2x/3), given x>0, x<`pi` ?

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Prove that `cosx<cos(x/3)cos((2x)/3)` for `0<x<pi` :

Note that on `0<x<pi` `sinx>0`

`cosx=cos(x/3+(2x)/3)` (Since `x/3+(2x)/3=x` )

`=cos( x/3)cos ((2x)/3)-sin (x/3)sin ((2x)/3)`

**Using `cos(A+B)=cosAcosB-sinAsinB`

`<cos(x/3)cos((2x)/3)` as required.

(We are subtracting a positive number)

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