x,y>0, x+y=1 tìm min,max
$\sqrt{x^{2014}+1} + \sqrt{y^{2014}+1}$
x,y>0 x+y=1 tìm min,max $\sqrt{x^{2014}+1} + \sqrt{y^{2014}+1}$
Started By ZzZzZzZzZ, 05-03-2014 - 21:41
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#2
Posted 06-03-2014 - 12:07
Kaito Kuroba
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Thiếu úy
x,y>0, x+y=1 tìm min,max
$\sqrt{x^{2014}+1} + \sqrt{y^{2014}+1}$
theo mincopki ta có:
$P\ge \sqrt{4+(\sum x^{1007})} $
mặt khác ta lại có:
$\sum x^{1007}\ge \frac{(x+y)^{1007}}{2^{1006}}$
Do đó $P\ge \sqrt{4+\frac{1}{2^{2012}}}$
$"="\Leftrightarrow $$x=y=\frac{1}{2}$
Edited by Kaito Kuroba, 06-03-2014 - 12:10.
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