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For fixed $g$, I want to find maximum $b$ with $$-2b(3t^2(s+1)+6t(s+1)+3s+2)-2g(6ts+3t+6s+2)-3ts^2+6ts+3t-3s^2+3s+1>0$$ for some nonnegative reals $t,s$. Here $g, b$ are also $\geq 0$. 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 $f(g)$.

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There might be some typo in the expression you want to be positive. You could check it. –  Did Aug 22 '11 at 20:49

1 Answer 1

up vote 3 down vote accepted

For every nonnegative $g$, $s$ and $t$, let $$ u(g,s,t)=\frac{-2g(6ts+3t+6s+2)-3ts^2+6ts+3t-3s^2+3s+1}{2(3t^2(s+1)+6t(s+1)+3s+2)}. $$ and $$ \varphi(g)=\inf\{u(g,s,t);s\ge0,t\ge0\}. $$ Then $f(g)=\varphi(g)$ fits your requirement and no number greater than $\varphi(g)$ can.

The problem is that for every $g$, $u(g,s,1)\to-\infty$ when $s\to+\infty$ hence $\varphi(g)=-\infty$ and no finite $b$ is such that $b<u(g,s,t)$ for every nonnegative $s$ and $t$.

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