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Ta cm bat dang thuc sau
$\sqrt{\frac{x^{2}+1}{2}}$ + $\sqrt{x}$ <= x+1.
Bien doi tuong duong ta co ($\sqrt{x}$ - 1)$^{4}$ >= 0.(dung).
Ap dung ta co $\sum \sqrt{\frac{a^{2}+1}{2}}$+$\sum \sqrt{a}$<=a+b+c+3.
theo bdt cosi ta co a+b+c>=3.
Tu hai bdt tren ta co dpcm
nguyenthehoanhsgs
#374861 $\sum \sqrt{a^{2}+1}\leq \sqrt...
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trong 03-12-2012 - 20:24
#374538 $\sum \frac{1}{x}\geq \sqrt...
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trong 02-12-2012 - 14:15
Cho a, b, c >0 thoa man a$^{2}$ + b$^{2}$ + c$^{2}$ = 3.
cmr
$\sum \frac{1}{x}\geq \sqrt{3\sum \frac{yz}{x}}$
cmr
$\sum \frac{1}{x}\geq \sqrt{3\sum \frac{yz}{x}}$
- nguyenthehoan yêu thích
#374410 Min P=$\frac{1}{x(x-1)^2}+\frac{1...
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trong 01-12-2012 - 22:58
ta cm minP = $\frac{1}{4}$
vi xy+yz+zx=xyz suy ra x, y, z >1 va $\sum \frac{1}{x}$=1
dat x=$\frac{b+c}{a}$+1, y=$\frac{c+a}{b}$+1, z=$\frac{a+b}{c}$+1 ta dua bdt can chung minh ve
$\sum \frac{a^{3}}{(a+b+c)(b+c)^{2}}$>=$\frac{1}{4}$
ta co $\sum \frac{a^{3}}{(b+c)^{2}}$>=$\frac{(a+b+c)^{3}}{4(a+b+c)^{2}}$=$\frac{a+b+c}{4}$ (bat dang thuc holder)
chia hai ve cho (a+b+c) ta co dpcm
vi xy+yz+zx=xyz suy ra x, y, z >1 va $\sum \frac{1}{x}$=1
dat x=$\frac{b+c}{a}$+1, y=$\frac{c+a}{b}$+1, z=$\frac{a+b}{c}$+1 ta dua bdt can chung minh ve
$\sum \frac{a^{3}}{(a+b+c)(b+c)^{2}}$>=$\frac{1}{4}$
ta co $\sum \frac{a^{3}}{(b+c)^{2}}$>=$\frac{(a+b+c)^{3}}{4(a+b+c)^{2}}$=$\frac{a+b+c}{4}$ (bat dang thuc holder)
chia hai ve cho (a+b+c) ta co dpcm
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