x+y+z=0

=>x+y=-z

=>y+z=-x

=>z+x=-y

(1+x/y)(1+y/z)(1+z/x)

(y+x/y)(z+y/z)(x+z/x)

-z/y.-x/z.-y/x

=-1

\(\text{Cho:}x^2+y^2+z^2=1\text{.Chứng minh rằng:}\frac{x^3}{y+2z}+\frac{y^3}{z+2x}+\frac{z^3}{z+2y}\ge\frac{1}{3}\)

\(\text{Áp dụng BĐT Cosi cho 2 số dương, ta có:}\)

\(\frac{9x^3}{y+2z}+x\left(y+2z\right)\ge6x^2;\frac{9y^3}{z+2x}+y\left(z+2x\right)\ge6y^2;\frac{9z^3}{x+2y}+z\left(x+2y\right)\ge6z^3\)

\(\text{Lại có:}\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\Rightarrow x^2+y^2+z^2\ge xy+yz+zx\)

\(\text{Do đó:}\frac{9x^3}{y+2z}+\frac{9y^3}{z+2x}+\frac{9z^3}{x+2y}+3\left(xy+yz+zx\right)\ge6\left(x^2+y^2+x^2\right)\)

\(\Leftrightarrow\frac{9x^3}{y+2z}+\frac{9y^3}{z+2x}+\frac{9z^3}{x+2y}\ge6\left(x^2+y^2+z^2\right)-3\left(xy+yz+zx\right)\ge3\left(x^2+y^2+z^2\right)\)

\(\Leftrightarrow\frac{x^3}{y+2z}+\frac{y^3}{z+2x}+\frac{z^3}{x+2y}\ge\frac{x^2+y^2+z^2}{3}=\frac{1}{3}\)

\(\text{Dấu "=" xảy ra }\Leftrightarrow x=y=z=\frac{1}{\sqrt{3}}\)

cho minh hoi phan bat dang thuc cosi la ban dung cong thuc the nao ak

ĐS: P=8

em cung ham mo tara

x³ + y³ + z³ =   3xyz

x³ + y³ + z³ – 3xyz = 0

x³ + y³ + z³ – xyz – xyz – xyz = 0

x³ + y³ + z³ – xyz – xyz – xyz  - x²y – y²x – x²z – z²x - y²z – z²y + x²y + y²x + x²z + z²x + y²z+ z²y = 0

(x³ + x²y + x²z) + (y³ + y²x + y²z) + (z³ + z²x + z²y) – ( xyz + x²y + y²x)  - (xyz + x²z + z²x) - (xyz +  z²y + y²z) = 0

( x + y + z ) ( x² + y² + z² – xy – xz – yz) = 0 ó ( x + y + z )( x – y)²(y – z)²(z – x)² = 0

=> x + y + z = 0  hoặc x = y = z 

mà theo đề ra thì x + y + z \(\ne\)0 nên x = y = z 

vậy P = ..............

x-y-z=0

=> x=y+z

     y=x-z

    -z=y-x

B=(1-z/x)(1-x/y)(1+y/z)

B=((x-z)/x)((y-x)/y)((z+y)/z)

B=(y/x)(-z/y)(x/z)

B=(-z.y.x)/(x.y.z)

B=-1

Áp dụng tính chất dãy tỉ số bằng nhau ta có

\(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}=\frac{a+b+c}{x+y+z}=\frac{1}{5}\)

\(\Rightarrow A=\frac{x+y+z}{a+b+c}=\frac{5}{1}=5\)

Vậy A = 5