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May 23, 2022 at 6:09 history edited Rodrigo de Azevedo CC BY-SA 4.0
added 33 characters in body; edited tags
Mar 5, 2013 at 13:12 comment added Barry Cipra I'm perplexed by what you're trying to maximize. As soon as you have one pair, $(u_0,v_0)$, that satisfies the inequality for all $x,y$, any pair $(u_0,v)$ with $v>v_0$ also satisfies the inequality. Letting $(u_0,v_0)=(1/2,1/2)$ (since $u_0x^2+v_0y^2-xy = (x-y)^2/2$), it looks like you wind up "maximizing" $2(1+v)$ over $v>1/2$.
Mar 5, 2013 at 1:57 vote accept Yuan
Mar 4, 2013 at 19:02 history edited Dima Pasechnik
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Mar 4, 2013 at 19:01 answer added Dima Pasechnik timeline score: 2
Mar 4, 2013 at 16:39 comment added Gilead It looks like a nonlinear fractional program (en.wikipedia.org/wiki/Fractional_programming). You can solve it as a general nonlinear program, but there may be specific properties you can exploit if you treat it as a fractional program.
Mar 4, 2013 at 16:21 history asked Yuan CC BY-SA 3.0