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Oct 27, 2020 at 17:30 comment added Iosif Pinelis @BPN : The minimization in $z$ can be done calculus-free, since $f(x,y,z)$ is quadratic in $z$. The case $x,y\le2$ is now also done in a calculus-free way. So, the entire proof is now calculus-free.
Oct 27, 2020 at 17:28 history edited Iosif Pinelis CC BY-SA 4.0
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Oct 27, 2020 at 17:23 comment added BPN If no elementary solutions materialise in a day or two, I'll happily accept this answer.
Oct 27, 2020 at 17:18 comment added BPN Many thanks to both you and Henri, I ended up bashing through a bit of calculus to arrive at precisely this eventually as well. However, it would be nice if there were an elementary proof of some sort: this inequality seems to be the first in a family of more complicated inequalities, and my reason for hoping for an "Olympiad-style" solution was that such an approach seems much more likely to generalise.
Oct 27, 2020 at 17:13 history answered Iosif Pinelis CC BY-SA 4.0