For any triangle ABC with sides a,b,c prove that a cos ⁡ A + b cos ⁡ B + c cos ⁡ C ≤ a + b + c 2 How to initiate this question?

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For any triangle ABC with sides a,b,c prove that  a  cos  ⁡  A  +  b  cos  ⁡  B  +  c  cos  ⁡  C  ≤            a      +      b      +      c        2  How to initiate this question?

kawiarkahh · Accepted Answer

If   a  ≥  b  ≥  c then   α  ≥  β  ≥  γ  and since cos is a decreasing function on   [  0  ,  π  ] , we obtain:  cos  ⁡  α  ≤  cos  ⁡  β  ≤  cos  ⁡  γ  .Thus, since  cos  ⁡  α  +  cos  ⁡  β  +  cos  ⁡  γ  =  1  +      r    R    ≤      3    2    ,by Chebyshov we obtain:      ∑          c      y      c        a  cos  ⁡  α  ≤      1    3    (  a  +  b  +  c  )  (  cos  ⁡  α  +  cos  ⁡  β  +  cos  ⁡  γ  )  ≤            a      +      b      +      c        2    .Done!

Esmeralda Lane · Accepted Answer

We need to prove that      ∑          c      y      c                  a      (              b        2            +              c        2            −              a        2            )              2      b      c        ≤            a      +      b      +      c        2  or      ∑          c      y      c            a    2    (      b    2    +      c    2    −      a    2    )  ≤      ∑          c      y      c            a    2    b  cor      ∑          c      y      c        (      a    4    −  2      a    2        b    2    +      a    2    b  c  )  ≥  0or      ∑          c      y      c        (      a    4    −      a    3    b  −      a    3    c  +      a    2    b  c  )  +      ∑          c      y      c        (      a    3    b  +      a    3    c  −  2      a    2        b    2    )  ≥  0  ,which is Schur and       ∑          c      y      c        a  b  (  a  −  b      )    2    ≥  0.Done!

For any triangle ABC with sides a,b,c prove that a cos ⁡ A + b

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