[Toán 10] Bất đẳng thức

T

tell_me_goobye

[TEX]x,y,z>0; \sum x=\sqrt{2}[/TEX]
[TEX]Min: \sqrt{(x+y)(y+z)(x+z)}. \sum \frac{\sqrt{x+y}}{z} [/TEX]

áp dụng holder ta sẽ CM BDT [TEX]\geq 4(x+y+z)[/TEX]

[TEX]( \sum \frac{\sqrt{y+z}}{x} )^2(\sum x(y+z))(\sum x(y+z)^2) \geq (\sum (y+z))^4 =16(\sum x)^4 [/TEX]
cần CM

[TEX](\sum x)^2(x+y)(y+z)(x+z) \geq 2 (\sum xy)(\sum x(y+z)^2) [/TEX]

có
[TEX](\sum x)^2 \geq 3(\sum xy) [/TEX]
tiếp tục

[TEX]3(x+y)(y+z)(x+z) \geq 2 (\sum x(y+z)^2) [/TEX]
[TEX] <=> \sum xy(x+y) \geq 6xyz [/TEX]
(đúng theo AM GM)
hoàn tất
từ đó lắp gt vào tìm dc min
 
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N

nhockthongay_girlkute

a,b,c >0

[TEX] \frac{a+\sqrt{ab}+\sqrt[3]{abc}}{3} \leq \sqrt[3]{a.\frac{a+b}{2}.\frac{a+b+c}{3}} [/TEX]

một bài mạnh hơn

[TEX] a+\sqrt[3]{ab.\frac{a+b}{2}} +\sqrt[3]{abc} \leq 3 \sqrt[3]{a.(\frac{a+b}{2}).( \frac{a+b+c}{3})}[/TEX]
 
L

legendismine

a,b,c >0

CM
[TEX] \sum \frac{(a-b-c)^2}{2a^2+(b+c)^2} \geq \frac{1}{2}[/TEX]
(b+c)22(b2+c2)cyc(abc)2a2+b2+c21(b+c)^2\le 2(b^2+c^2)\Leftrightarrow\sum_{cyc}\frac {(a-b-c)^2}{a^2+b^2+c^2}\ge 1...........
(abc)2a2+b2+c213!!!!!!!!\frac {(a-b-c)^2}{a^2+b^2+c^2}\ge \frac {1}{3}!!!!!!!!
 
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L

legendismine

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N

nhockthongay_girlkute

cho x,y,z >0 x+y+z=1

CM
[TEX] \sum \frac{1}{yz+x+\frac{1}{x}} \leq \frac{27}{31} [/TEX]

( SERBIAN NATIONAL OLYMPIAD 2008 )
 
N

nhockthongay_girlkute

CHO a,b,c >0

CM

[TEX] \frac{a^2+b^2+c^2}{ab+bc+ac} +\frac{3}{2} . \frac{(a^2b+b^2c+c^2a)-abc}{(ab^2+bc^2+ca^2)-abc} \geq \frac{5}{2}[/TEX]
 
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