[Toán 9]Bất đẳng thức

B

bimkutepro11

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C

congchuaanhsang

Đặt a=1x\dfrac{1}{x} ; b=1y\dfrac{1}{y} ; c=1z\dfrac{1}{z}\Rightarrowx,y,z>0 và xyz=1
Gọi A là vế trái của BĐT
A=11x3(1y+1z)\dfrac{1}{\dfrac{1}{x^3}(\dfrac{1}{y}+\dfrac{1}{z})}+11y3(1x+1z)\dfrac{1}{\dfrac{1}{y^3}(\dfrac{1}{x}+ \dfrac{1}{z} )}+11z3(1x+1y)\dfrac{1}{\dfrac{1}{z^3}(\dfrac{1}{x}+ \dfrac{1}{y}) }
\LeftrightarrowA=x2y+z\dfrac{x^2}{y+z}+y2x+z\dfrac{y^2}{x+z}+z2x+y\dfrac{z^2}{x+y}
Áp dụng BĐT Cauchy - Schwarz
A=x2y+z\dfrac{x^2}{y+z}+y2x+z\dfrac{y^2}{x+z}+z2x+y\dfrac{z^2}{x+y}\geq(x+y+z)22(x+y+z)\dfrac{(x+y+z)^2}{2(x+y+z)}=x+y+z2\dfrac{x+y+z}{2}
Áp dụng BĐT Cauchy: x+y+z\geq3xyz3\sqrt[3]{xyz}=3\Rightarrowx+y+z2\dfrac{x+y+z}{2}\geq32\dfrac{3}{2}
\RightarrowA\geq32\dfrac{3}{2}
Dấu "=" xảy ra \Leftrightarrow a=b=c=1
 
C

congchuaanhsang

Đây bạn nhé:

1x3(1y+1z)\dfrac{1}{x^3(\dfrac{1}{y}+\dfrac{1}{z})}=1y+zx3yz\dfrac{1}{\dfrac{y+z}{x^3yz}}

xyz=1\Rightarrowx3yzx^3yz=x2x^2

\Rightarrow1y+zx3yz\dfrac{1}{\dfrac{y+z}{x^3yz}}=1y+zx2\dfrac{1}{\dfrac{y+z}{x^2}}=x2y+z\dfrac{x^2}{y+z}
 
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