If 1 2 + a + 1 2 + b + 1 2 + c = 1, prove a b + a c + b c ≤ 3

Question

If       1          2      +      a        +      1          2      +      b        +      1          2      +      c        =  1,  prove       a    b    +      a    c    +      b    c    ≤  3

skinutimanurfh · Accepted Answer

The condition gives:      3    2    −  1  =      ∑          c      y      c            (          1      2        −          1              2        +        a              )      ⟹    1  =      ∑          c      y      c            a          2      +      a        ≥            (              a            +              b            +              c                    )        2                    a      +      b      +      c      +      6          by Cauchy-Schwarz inequality    ⟹    3  ≥      a    b    +      b    c    +      c    a

Jamison Estrada · Accepted Answer

The condition gives that there are    α  ≥  0,   β  ≥  0 and   γ  ≥  0 such that   α  +  β  +  γ  =  π for which       a    b    =  2  cos  ⁡  γ,       a    c    =  2  cos  ⁡  β and       b    c    =  2  cos  ⁡  αHence, we need to prove that   cos  ⁡  α  +  cos  ⁡  β  +  cos  ⁡  γ  ≤      3    2    which is obvious.

If 1 2 + a </mrow> </mfrac> + 1

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