For fixed g , I want to find maximum b with &#x2212;<!-- − --> 2 b ( 3

Leland Morrow

Leland Morrow

Answered question

2022-06-06

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 > 0
for 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 < f ( g )? If one can find t, s for which b is maximum, one will get f ( g ).

Answer & Explanation

Belen Bentley

Belen Bentley

Beginner2022-06-07Added 28 answers

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 < u ( g , s , t ) for every nonnegative s and t.

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