Let α, β and γ be the three angles in a non-right triangle. How can i prove the following inequality? 1 + sin 2 ⁡ ( α ) cos 2 ⁡ ( α ) + 1 + sin 2 ⁡ ( β ) cos 2 ⁡ ( β ) + 1 + sin 2 ⁡ ( γ ) cos 2 ⁡ ( γ ) ≥ 1 + sin ⁡ ( α ) sin ⁡ ( β ) 1 − sin ⁡ ( α ) sin ⁡ ( β ) + 1 + sin ⁡ ( β ) sin ⁡ ( γ ) 1 − sin ⁡ ( β ) sin ⁡ ( γ ) + 1 + sin ⁡ ( α ) sin ⁡ ( γ ) 1 − sin ⁡ ( α ) sin ⁡ ( γ )

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Let   α,   β and   γ be the three angles in a non-right triangle. How can i prove the following inequality?                    1        +                  sin          2                ⁡        (        α        )                              cos          2                ⁡        (        α        )              +                    1        +                  sin          2                ⁡        (        β        )                              cos          2                ⁡        (        β        )              +                    1        +                  sin          2                ⁡        (        γ        )                              cos          2                ⁡        (        γ        )              ≥                    1        +        sin        ⁡        (        α        )        sin        ⁡        (        β        )                    1        −        sin        ⁡        (        α        )        sin        ⁡        (        β        )              +                    1        +        sin        ⁡        (        β        )        sin        ⁡        (        γ        )                    1        −        sin        ⁡        (        β        )        sin        ⁡        (        γ        )              +                    1        +        sin        ⁡        (        α        )        sin        ⁡        (        γ        )                    1        −        sin        ⁡        (        α        )        sin        ⁡        (        γ        )

lorienoldf7 · Accepted Answer

∑          c      y      c                  (      2      b      c              )        2            +      16              Δ        2                    (              b        2            +              c        2            −              a        2                    )        2              ≥      ∑          c      y      c                  4              R        2            +      a      b              4              R        2            −      a      b      is equivalent to  16      Δ    2        ∑          c      y      c            1          (              b        2            +              c        2            −              a        2                    )        2              ≥      ∑          c      y      c                  a      b              4              R        2            −      a      b      or to      ∑          c      y      c            1          (              b        2            +              c        2            −              a        2                    )        2              ≥      ∑          c      y      c            1          (              b        2            +              c        2            −              a        2                    )        2            +      4      (      a      −      b      )      b              c        2            which can be proved by bashing. Although I suspect there are more efficient ways, based on the convexity of   f  (  x  )  =            1      +      x              1      −      x       over   (  −  1  ,  1  ).

Let α , β and γ be the three ang

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