where is a? Let a,b.........

April wrote: Let $ \alpha$, $ \beta$ be acute angles. Find the maximum value of \[ \frac{\left(1-\sqrt{\tan\alpha\tan\beta}\right)^{2}}{\cot\alpha\cot\beta}. \] Something is wrong here. If we take $ x^{2}=\tan\alpha\tan\beta$ then $ \frac{\left(1-\sqrt{\tan\alpha\tan\beta}\right)^{2}}{\cot\alpha\cot\beta}=(x(x-1))^{2}$ and since $ \tan\alpha \in (0;\infty)$ with $ \alpha \in \left(0;\frac{\pi}{2}\right)$ so given expression doesn't have maximum value. In pdf file here: http://www.mathlinks.ro/Forum/viewtopic.php?t=161544 it is given different form of the expression: $ \frac{\left(1-\sqrt{\tan\alpha\tan\beta}\right)^{2}}{\cot\alpha+\cot\beta}$. However this also doesn't have maximum value with given conditions.

hello, i think it is $ \frac{\left(1-\sqrt{\tan(\alpha)\tan(\beta)}\right)^{2}}{\cot(\alpha)+\cot(\beta)}\le \frac{2}{27}$. Sonnhard.

April wrote: Let $ \alpha$, $ \beta$ be acute angles. Find the maximum value of \[ \frac {\left(1 - \sqrt {\tan\alpha\tan\beta}\right)^{2}}{\cot\alpha + \cot\beta} \] Let $ x=\tan \alpha$ $ y=\tan \beta$ Then $ P=\frac{xy (\sqrt{xy}-1)^2)}{x+y}$ Let $ x=y$ $ P=\frac{x(x-1)^2}{2}$ If $ x\to\infty$ then $ p\to \infty$ Imply that $ P$ has not max value.

Maybe $ \alpha +\beta$is constant ?!

maybe we need minimal value?

Extremal wrote: maybe we need minimal value? I don't think that because $ \tex{min} P = 0$ when $ xy = 1\Longleftrightarrow \alpha + \beta = \frac {\pi}{2}$

click on the first pdf: https://www.google.com/search?q=Chinese+Northern+Mathematical+Olympiad+2007&rlz=1C1HIJA_enUS786US786&oq=Chine&aqs=chrome.0.69i59j0l2j69i57j0l2.844j0j7&sourceid=chrome&ie=UTF-8 the problem is on the second page

https://artofproblemsolving.com/downloads/printable_post_collections/3577

Wait so does anyone know what the problem should be, cause I also got that it can approach infinity

It should be determine the maximum value of $ \frac{\left(1-\sqrt{\tan\frac{\alpha}{2}\tan\frac{\beta}{2}}\right)^{2}}{\cot\alpha+\cot\beta}$.