Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $f(g)$.

share|cite|improve this question
There might be some typo in the expression you want to be positive. You could check it. – Did Aug 22 '11 at 20:49
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 $t$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.