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I have function $$f(y)=\frac{1}{2} y \log \left(\frac{a^2 b \left(\frac{2}{y}-2\right)}{a b \left(\frac{2}{y}-2\right)+a+1}+1\right)$$ where $a,b>0$ and $y\in[0,1]$.

I want to show that $f(y)$ has a maximum at $y_{\max}$ in $y\in[0,1]$. If possible, it is better to find $y_{\max}$

I get $$f^{''}(y)=\frac{2 a^3 b^2 (a (4 b (y-1)-y)-2 y)}{(y-2 a b (y-1))^2 (a (y-2 b (y-1))+y)^2}.$$ Since $2 a^3 b^2>0$ and $(y-2 a b (y-1))^2 (a (y-2 b (y-1))+y)^2>0$, in order to have $f^{''}(y)<0$, I should have $a (4 b (y-1)-y)-2 y<0$, which can be shown as $a (4 b (y-1)-y)-2 y=-(a (4 b (1-y)+y)+2 y)<0$ since $y\in[0,1]$

1) Is this analysis enough to prove that $f(y)$ has only one maximum in $[0,1]$?

Further,

2) Does any one have an idea of deriving $y_{\max}$ analytically? I found that it is difficult as $f^{'}(y)=0$ in complected form: $$\frac{1}{2} \left(\log \left(1-\frac{2 a^2 b (y-1)}{a (-2 b y+2 b+y)+y}\right)-\frac{2 a^2 b y}{(a (2 b (y-1)-y)-y) (2 a b (y-1)-y)}\right)=0$$

Gerry Myerson
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Frey
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1 Answers1

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This is not an answer but it is probably too long for a comment.

I think that Chinny84 made a very good point since you have $$f(0)=0$$ $$f(1)=0$$ $$f'(0)=\frac{1}{2} \log (a+1)>0$$ $$f'(1)=-\frac{a^2 b}{a+1}<0$$ $$f''(0)=-\frac{1}{2 b}$$ $$f''(1)=-\frac{2 a^3 (a+2) b^2}{(a+1)^2}$$ and you have shown that $f''(y)<0$ for any $y$ in the interval.

Trying to solve $f'(y)=0$ seems to be close to impossible (at least to me). I suppose that it could be somehow possible to make approximation $$f(y)\approx y\,(1-y)\,g(y)$$ and use the above specific values to adjust parameters in an appropriate function $g(y)$. I must confess that, for the general case, I did not find anything satisfactory.

Claude Leibovici
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