Chứng minh rằng :
$\frac{a + b}{2} \leq \sqrt{\frac{a^{2} + b^{2}}{2}}$.
$\frac{a + b}{2} \leq \sqrt{\frac{a^{2} + b^{2}}{2}}$
Started By tramyvodoi, 12-02-2013 - 22:30
#1
Posted 12-02-2013 - 22:30
#2
Posted 12-02-2013 - 22:33
sao cứ post đi post lại 1 bài mãi thế chị?Chứng minh rằng :
$\frac{a + b}{2} \leq \sqrt{\frac{a^{2} + b^{2}}{2}}$.
C-S ( để dấu giá trị tuyệt đối)=> dpcm
- giacatluongpro1997 likes this
#3
Posted 09-03-2013 - 22:24
Từ BĐT $(a+b)^{2}\leq 2(a^{2}+b^{2})\Rightarrow \frac{(a+b)^{2}}{4}\leq \frac{a^{2}+b^{2}}{2}\Rightarrow \frac{a+b}{2}\leq \sqrt{\frac{a^{2}+b^{2}}{2}}$
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Chuyên Vĩnh Phúc
#4
Posted 18-03-2013 - 16:13
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