Show that for real a , b , c, a 2 + b 2 + c 2 ≥ a b + b c + c a

Question

Show that for real   a  ,  b  ,  c,       a    2    +      b    2    +      c    2    ≥  a  b  +  b  c  +  c  a

Zackery Harvey · Accepted Answer

Let   f  (  a  ,  b  ,  c  )  =      a    2    +      b    2    +      c    2    −  a  b  −  b  c  −  c  a. Then we have,   f  (  t  a  ,  t  ,  b  ,  t  c  )  =      t    2    f  (  a  ,  b  ,  c  ). Hence,   f is homogeneous of degree two. For,   t  ≠  0, we have   f  (  a  ,  b  ,  c  )  &amp;gt;  0    ⟺    f  (  t  a  ,  t  b  ,  t  c  )  &amp;gt;  0. Therefore, we may make various normalizations. For example, we may set,   a  =  1  ,  b  =  1  +  x  ,  c  =  1  +  y and get       x    2    +      y    2    −  x  y  =  (  x  −      y    2        )    2    +                    3                  y          2                    4        &amp;gt;  0

malalawak44 · Accepted Answer

Using caushy-Schwartz Inequality:      (          a      2        +          b      2        +          c      2        )    ⋅    (          b      2        +          c      2        +          a      2        )    ≥    (    a    b    +    b    c    +    c    a          )      2      and equality hold when             a      b        =          b      c        =          c      a        .

Show that for real a , b , c , a 2 </msup> + b 2

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