Is there a real solution to: B ≤ A , B ≤ C? Namely, is there s = ( α , β , γ , δ ) ∈ R 4 such that B ( s ) ≤ A ( s ) , B ( s ) ≤ C ( s )?

Question

Is there a real solution to:   B  ≤  A  ,  B  ≤  C? Namely, is there   s  =  (  α  ,  β  ,  γ  ,  δ  )  ∈        R    4   such that   B  (  s  )  ≤  A  (  s  )  ,  B  (  s  )  ≤  C  (  s  )?

Sanaa Hinton · Accepted Answer

Let       A    ~    (  s  )  =  A  (  s  )  −  B  (  s  ) and       C    ~    (  s  )  =  C  (  s  )  −  B  (  s  )Since   A  ,  B  ,  C are homogeneous of degree   3, so do       A    ~   and       C    ~  . For any   s  ∈        R    4  , we have       A    ~    (  −  s  )  =  −      A    ~    (  s  ) and       C    ~    (  −  s  )  =  −      C    ~    (  s  ).Using continuity of       A    ~  , we can find a point   t  ∈        R    4   with   |  t  |  =  1 and       A    ~    (  t  )  =      A    ~    (  −  t  )  =  0. Since at least one of       C    ~    (  t  ) or       C    ~    (  −  t  )  ≥  0. We have find a   s  =  t or   −  t such that   |  s  |  =  1 and      A    ~    (  s  )  =  0  ,      C    ~    (  s  )  ≥  0      ⟺      A  (  s  )  =  B  (  s  )  ,  C  (  s  )  ≥  B  (  s  )

Is there a real solution to: B ≤ A , B ≤ C

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