Find the number of solutions to a − b 2 ≥ 1 4 , b − c 2 ≥ 1 4 , c − d 2 ≥ 1 4 , d − a 2 ≥ 1 4

Question

Find the number of solutions to  a  −      b    2    ≥      1    4    ,  b  −      c    2    ≥      1    4    ,  c  −      d    2    ≥      1    4    ,  d  −      a    2    ≥      1    4

elladanzaez · Accepted Answer

Adding all 4 equations,  a  +  b  +  c  +  d  −  (      a    2    +      b    2    +      c    2    +      d    2    )  ≥  1  a  (  1  −  a  )  +  b  (  1  −  b  )  +  c  (  1  −  c  )  +  d  (  1  −  d  )  ≥  1Using symmetry,  a  (  1  −  a  )  ≥            1      4        b  (  1  −  b  )  ≥            1      4        c  (  1  −  c  )  ≥            1      4        d  (  1  −  d  )  ≥            1      4      Considering   a  (  1  −  a  )  ≥            1      4      ,  4  a  −  4      a    2    ≥  1  ⇒  4      a    2    −  4  a  +  1  ≤  0  ⇒  (  2  a  −  1      )    2    ≤  0We can Add all these as  4  a  −  4      b    2    +  4  b  −  4      c    2    +  4  c  −  4      d    2    +  4  d  −  4      a    2    ≥  1  +  1  +  1  +  1So  (  4      a    2    −  4  a  +  1  )  +  (  4      b    2    −  4  b  +  1  )  +  (  4      c    2    −  4  c  +  1  )  +  (  4      d    2    −  4  d  +  1  )  ≤  0So  (  2  a  −  1      )    2    +  (  2  b  −  1      )    2    +  (  2  c  −  1      )    2    +  (  2  d  −  1      )    2    ≤  0So we have  (  2  a  −  1      )    2    +  (  2  b  −  1      )    2    +  (  2  c  −  1      )    2    +  (  2  d  −  1      )    2    =  0So  a  =  b  =  c  =  d  =      1    2

Find the number of solutions to a − b 2 </msup> &#x2265

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