Question

[Vũ Quang Huy]

Cho x^2 +y^2+z^2 =1 va x,y,z > 0 Chứng minh x^3/(y+2z)+y^3/(z+2x)+z^3/(x+2y)>=1/3

Answer 1

[✧ ( •̀ ω •́ )✧₸ĦV✧( •̀ ω •́ )✧]

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

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

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

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

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

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

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

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

Answer 2

[Vũ Quang Huy]

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