For fixed g, I want to find maximum b with − 2 b ( 3 t 2 ( s + 1 ) + 6 t ( s + 1 ) + 3 s + 2 ) − 2 g ( 6 t s + 3 t + 6 s + 2 ) − 3 t s 2 + 6 t s + 3 t − 3 s 2 + 3 s + 1 &gt; 0for some nonnegative reals t, s. Here g, b are also ≥ 0. Can it be possible to get a function f such that we get the upper bound on b as f ( g ) i.e, b &lt; f ( g )? If one can find t, s for which b is maximum, one will get f ( g ).

Question

For fixed   g, I want to find maximum   b with  −  2  b  (  3      t    2    (  s  +  1  )  +  6  t  (  s  +  1  )  +  3  s  +  2  )  −  2  g  (  6  t  s  +  3  t  +  6  s  +  2  )  −  3  t      s    2    +  6  t  s  +  3  t  −  3      s    2    +  3  s  +  1  &amp;gt;  0for some nonnegative reals   t,  s. Here   g,  b are also   ≥  0. Can it be possible to get a function   f such that we get the upper bound on   b as   f  (  g  ) i.e,   b  &amp;lt;  f  (  g  )? If one can find   t,  s for which   b is maximum, one will get   f  (  g  ).

Belen Bentley · Accepted Answer

For every nonnegative   g,   s and   t, let  u  (  g  ,  s  ,  t  )  =            −      2      g      (      6      t      s      +      3      t      +      6      s      +      2      )      −      3      t              s        2            +      6      t      s      +      3      t      −      3              s        2            +      3      s      +      1              2      (      3              t        2            (      s      +      1      )      +      6      t      (      s      +      1      )      +      3      s      +      2      )        .and  φ  (  g  )  =  inf  {  u  (  g  ,  s  ,  t  )  ;  s  ≥  0  ,  t  ≥  0  }  .Then   f  (  g  )  =  φ  (  g  ) fits your requirement and no number greater than   φ  (  g  ) can.The problem is that for every   g,   u  (  g  ,  s  ,  1  )  →  −  ∞ when   s  →  +  ∞ hence   φ  (  g  )  =  −  ∞ and no finite   b is such that   b  &amp;lt;  u  (  g  ,  s  ,  t  ) for every nonnegative   s and   t.

For fixed g , I want to find maximum b with − 2 b ( 3

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