Bài 1 : Cho $ x^{2}+4y^{2}=1$ .Chứng minh$ \left | x-y \right |\leq \dfrac{\sqrt{5}}{2}$
Chứng minh bất đẳng thức
Started By chit_in, 31-10-2011 - 23:16
#2
Posted 01-11-2011 - 09:20
PRONOOBCHICKENHANDSOME
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$(x-y)^2 = (1.x-\dfrac{1}{2}.2.y)^2 \overset {C-S} {\le} (1 +\dfrac{1}{4})(x^2+4y^2)=\dfrac{5}{4}$
$\Rightarrow |x-y| \leq \dfrac{\sqrt{5}}{2}$
$\Rightarrow |x-y| \leq \dfrac{\sqrt{5}}{2}$
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