Function Analysis: Proving h'(y_s) < 0 for All y_s

[standardflop]

Hello,

given is the function h(y_s) = \ln (1-y_s) - \ln y_s - \gamma + \frac{\gamma}{\theta + \beta (1-y_s)}
my job is now to show that h&#039;(y_s) &lt; 0, \forall y_s \in ]0,1[ when
\frac{\gamma \beta}{\theta (\beta + \theta)} &lt; 4

I guess that all constants can be assumed to be real and positive.
My first tought was to introduce the new variable z=1-y_s so that after differentiation i would get

- 1/z + 1/(z-1) + \frac{\gamma \beta}{(\theta + \beta z)^2} &lt; 0
but i can't derive the above expression from this inequality. Any help would be greatly appreciated.

Thank you.

 

[HallsofIvy]

Since 0&lt; y_s&lt; 1[/tex], then 0&amp;lt; z&amp;lt; 1<br /> Adding the first two terms, what you have is<br /> \frac{-1}{z(z-1)}+\frac{\gamma\beta}{(\theta+ \beta z)^2}<br /> Now what is the <b>smallest</b> that first term could be? What is the <b>largest</b> the second term could be? (Remember the condition that <br /> \frac{\gamma \beta}{\theta (\beta + \theta)} &amp;lt; 4).

 

[standardflop]

The second term is largest when z \rightarrow 0, where it takes the values \frac{\gamma \beta}{\theta^2}. But as i see it, the first term is a problem since it goes toward \pm \infty (or undef.?) when z \rightarrow 0 \ \wedge \ z \rightarrow 1 respectively.? So can this term be bounded in the interval 0&lt;z&lt;1 ?

Also i believe your first term has the wrong sign.

 

[HallsofIvy]

Sorry about the sign. When I did it I was thinking 1/(1-z)- 1/z but then used your "reversed" z-1. Yes, it is
\frac{1}{z(z-1)}+\frac{\gamma\beta}{(\theta+ \beta z)^2}
or, what I really intended,
\frac{-1}{z(1-z)}+\frac{\gamma\beta}{(\theta+ \beta z)^2}
That first term, whether 1/(z(z-1)) or -1/(z(1-z)) is always negative. And when I said "smallest value", I should have made it clear that I was thinking of the absolute value because I was focusing on the subtraction- the largest possible value of the second term, minus the smallest possible value of the first is what?

 

[standardflop]

The largest possible value of the second term, minus the (abs.) smallest possible value of the first is
\frac{\gamma \beta}{\theta^2} - 4 &lt; 0 because \vline \ \max_{0&lt;z&lt;1} \ \frac{1}{z(z-1)} \ \vline = 4

which leads to \frac{\gamma \beta}{\theta^2} &lt; 4 and not the expected \frac{\gamma \beta}{\theta (\beta + \theta)} &lt; 4

 

[standardflop]

=> or is this to be understood as a less strict requirement, because \frac{\gamma \beta}{\theta (\beta + \theta)} &lt; \frac{\gamma \beta}{\theta^2} ? Therefore if \frac{\gamma \beta}{\theta^2} &lt; 4 then also \frac{\gamma \beta}{\theta (\beta + \theta)} &lt; 4 ?