Find min of M = 1 2 + cos ⁡ 2 A + 1 2 + cos ⁡ 2 B + 1 2 − cos ⁡ 2 C , where A,B,C are three angle of triangle ABC

Question

Find min of  M  =      1          2      +      cos      ⁡      2      A        +      1          2      +      cos      ⁡      2      B        +      1          2      −      cos      ⁡      2      C      , where A,B,C are three angle of triangle ABC

Arianna Turner · Accepted Answer

Using Cauchy-Schwarz, we obtain:                    M                            =                  1                      2            +            cos            ⁡            2            A                          +                  1                      2            +            cos            ⁡            2            B                          +                  1                      2            −            cos            ⁡            2            C                                                            ≥                  9                      6            +            2            cos            ⁡            (            A            +            B            )            cos            ⁡            (            A            −            B            )            +            1            −            2                          cos              2                        ⁡            C                                                            ≥                  9                      7            −            2            cos            ⁡            C            −            2                          cos              2                        ⁡            C                          ≥                  9                      15            2                          =                  6          5                    After using C-S we need to prove that                                                9                          6              −              2              cos              ⁡              γ              cos              ⁡                              (                α                −                β                )                            +              1              −              2                              cos                2                            ⁡              γ                                ≥                      6            5                                                      ⟺                                      4                      cos            2                    ⁡          γ          +          4          cos          ⁡          γ          cos          ⁡                      (            α            −            β            )                    +          1          ≥          0                                            ⟺                                                                (              2              cos              ⁡              γ              +              cos              ⁡                              (                α                −                β                )                            )                        2                    +                      sin            2                    ⁡                      (            α            −            β            )                    ≥          0                    Done!

Alaina Marshall · Accepted Answer

The change might not be correct, but before that the numerator is  f  (  A  ,  B  )  =  6  +  cos  ⁡  2  A  +  cos  ⁡  2  B  −  cos  ⁡  2  C  =  6  +  cos  ⁡  2  A  +  cos  ⁡  2  B  −  cos  ⁡  2  (  A  +  B  )            ∂      f              ∂      A        =  2  sin  ⁡  (  2  A  +  2  B  )  −  2  sin  ⁡  (  2  A  )            ∂      f              ∂      B        =  2  sin  ⁡  (  2  A  +  2  B  )  −  2  sin  ⁡  (  2  B  )We can only have extrema at the border of the set, namely A=0, B=0,   A  +  B  =  π or where both partials are 0, meaning   sin  ⁡  2  A  =  sin  ⁡  2  B, equivalently A=B or   A  +  B  =  k      π    2  . From checking these we get that the minimum is at       π    6    ,      π    6   and is in fact       6    5

Find min of M = 1 2 + cos ⁡

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