Prove the next cyclic inequalityLet a , b , c &gt; 0 such that a + b + c = 1prove that c y c l i c ∑ a b c + a b ≤ 1 2 By symetrie, i proved it by assuming a = b = c but i cannot justify this hypothesis.

Question

Prove the next cyclic inequalityLet   a  ,  b  ,  c  &amp;gt;  0 such that  a  +  b  +  c  =  1prove that  c  y  c  l  i  c  ∑            a      b              c      +      a      b        ≤      1    2  By symetrie, i proved it by assuming   a  =  b  =  c but i cannot justify this hypothesis.

Milton Gilmore · Accepted Answer

By C-S            (              ∑                  c          y          c                                      a          b                          c          +          a          b                    )        2    =            (              ∑                  c          y          c                                      a          b                          c          (          a          +          b          +          c          )          +          a          b                    )        2    =            (              ∑                  c          y          c                                      a          b                          (          a          +          c          )          (          b          +          c          )                    )        2    ≤  ≤      ∑          c      y      c        a  b      ∑          c      y      c                  a      b              (      a      +      c      )      (      b      +      c      )        .Thus, it&#039;s enough to prove that      ∑          c      y      c        a  b      ∑          c      y      c                  a      b              (      a      +      c      )      (      b      +      c      )        ≤            (      a      +      b      +      c              )        2              4    ,which is easy after full expanding.Indeed, we need to prove that      ∑          c      y      c        (      a    4    b  +      a    4    c  −      a    3        b    2    −      a    3        c    2    −  2      a    3    b  c  +  2      a    2        b    2    c  )  ≥  0or      ∑          c      y      c        (  a  −  b      )    2    a  b  (  a  +  b  −  c  )  ≥  0.Let   a  ≥  b  ≥  cThus,      ∑          c      y      c        (  a  −  b      )    2    a  b  (  a  +  b  −  c  )  ≥  ≥  (  a  −  c      )    2    a  c  (  a  +  c  −  b  )  +  (  b  −  c      )    2    b  c  (  b  +  c  −  a  )  ≥  ≥  (  b  −  c      )    2    a  c  (  a  −  b  )  +  (  b  −  c      )    2    b  c  (  b  −  a  )  =  (  b  −  c      )    2    (  a  −  b      )    2    c  ≥  0.Done!

Prove the next cyclic inequality Let a,b,c>0 such that a+b+c=1 prove that cyclic sum (ab)/(sqrt (c+ab))<=1/2 By symetrie, i proved it by assuming a=b=c but i cannot justify this hypothesis.

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