I have two sets of inequalities and i just want to know if they are correct.

The parameters $\mu, K, d_1, \sigma_1,\sigma_2$ and dependent variables $H,F$are positive. Also $\sigma_2>\sigma_1$.

\begin{align*} \frac{dH}{dt}&=\mu \frac{(H+F)^2}{K^2+(H+F)^2}-d_1H-H\left(\sigma_1-\frac{\sigma_2 F}{H+F}\right)\\ &\leq \mu-(d_1+\sigma_1-\sigma_2)H\\ &\leq 0 \quad \textrm{if}\quad H\geq \frac{\mu}{d_1+\sigma_1-\sigma_2} \end{align*}

Also this one:

\begin{align*} \frac{dF}{dt}&=H\left(\sigma_1-\frac{\sigma_2 F}{H+F}\right)-(p+d_2)F \\ &\leq H\left(\sigma_1-\frac{\sigma_2 F}{H+F}\right)\\ &\leq (\sigma_1-\sigma_2)H\\ &<0 \end{align*}

  • 1
    $\begingroup$ To see the first inequality use that $\frac{a}{a+b}<1$ for $a,b>0.$ with respect to the second one, what assumptions have you on $p,d_2?$ $\endgroup$ – mfl Jun 27 '14 at 20:02
  • $\begingroup$ $p$ and $d_2$ are both positive. So I left them and put a $\leq$ sign. Do the first set of inequalities make sense? $\endgroup$ – math Jun 27 '14 at 20:37
  • $\begingroup$ The first set of inequalities is correct. In the second one, what happens to $a+b$ if $b<0?$ Is it $a+b<b?$ Do you see it? $\endgroup$ – mfl Jun 27 '14 at 20:40
  • $\begingroup$ By $a$ and $b$ you mean $p$ and $d_2$? $\endgroup$ – math Jun 27 '14 at 20:42
  • 1
    $\begingroup$ No, if $p>0,d_2>0,F>0$ then $-(p+d_2)F<0.$ So $\frac{dF}{dt}=\mathrm{something}-(p+d_2)F<\mathrm{something}.$ $\endgroup$ – mfl Jun 27 '14 at 20:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.