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Define function $$f(\alpha)=\frac{-(4-5\alpha-4\alpha^{2})+\sqrt{32\alpha^{3}+\alpha^{2}-40\alpha+16}}{(1-2\alpha)\alpha}$$ where $\alpha \in [0,\frac{1}{2}]$ I can verify that:

(1) $f(\alpha)\geq 0$ always holds when $\alpha \in [0,\frac{1}{2}]$

(2) Using L'Hôpital's rule to show $f(0)=0$ and $f(\frac{1}{2})=6$

I can plot this function via MATLAB and observes this function is increasing when $\alpha \in [0,\frac{1}{2}]$.

My goal is to prove $f(\alpha)$ is increasing in $\alpha \in [0,\frac{1}{2}]$, the geometric graph is simple but the verification of increasing is not easy. How to show it?

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    $\begingroup$ This doesn't seem to be a research level problem in mathematics. Please move it to the MSE because MSE is more relevant for such questions. $\endgroup$ Commented Aug 10, 2017 at 15:56
  • $\begingroup$ @adityaguharoy, when $\alpha=0.5$, the numerator is also 0, I use LHopital rule to calculate it and get $f(\alpha=0.5)=6$ $\endgroup$
    – Galor
    Commented Aug 10, 2017 at 16:30
  • $\begingroup$ @adityaguharoy, thanks for notice, I will move it. $\endgroup$
    – Galor
    Commented Aug 10, 2017 at 16:30
  • $\begingroup$ L-Hopital rule doesn't say that we can take derivatives of $\frac{f}{g}$ at a point $x$ where both $f(x)=0$ and $g(x)=0$. In fact we must ensure that the denominator is always $\ne0$ . Once you get a $0$ denominator the function isn't defined anymore (taking values in $\mathbb{R}$) $\endgroup$ Commented Aug 10, 2017 at 16:33
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    $\begingroup$ @adityaguharoy, yes this is an elementary question, I should post in MSE. The background of this problem guarantees that alpha's domain belongs in (0,0.5), while alpha only tends to 0 or 0.5 in limiting situation(alpha is a function of another x defined in R, alpha tends to 0 or 0.5 when x tends to positive or negative infinity). Then here, it is easy to verify I can apply LHopital rule here. As when alpha in (0,0.5), both f and g are not 0. as alpha tends to 0.5, both f, g tends to 0, while f' and g' exist and are not zero at point 0.5 $\endgroup$
    – Galor
    Commented Aug 10, 2017 at 16:46

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