Question

[Vũ Anh Quân]

Cho \(P=\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\)

và \(Q=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\)

Chứng minh nếu P=1 thì Q=0

Answer 1

[Đức Cường]

Bài này mình làm 2 cách cho bạn dễ hiểu nha

C1:\(P=\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}=1\Leftrightarrow x\left(z+x\right)\left(x+y\right)+y\left(y+z\right)\left(x+y\right)+z\left(z+x\right)\left(y+z\right)=\left(y+z\right)\left(x+y\right)\left(z+x\right) \)\(\Leftrightarrow x^2\left(y+z\right)+y^2\left(x+z\right)+z^2\left(x+y\right)+x^3+y^3+z^3+3xyz=x^2\left(y+z\right)+y^2\left(x+z\right)+z^2\left(x+y\right)+2xyz\)

\(\Leftrightarrow x^3+y^3+z^3+xyz=0\)

\(\Rightarrow\left(x^3+y^3+z^3+xyz\right)\left(x+y+z\right)=0 \)

Ta cũng thấy Q=\(Q=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}=\dfrac{x^2\left(z+x\right)\left(x+y\right)+y^2\left(y+z\right)\left(x+y\right)+z^2\left(y+z\right)\left(z+x\right)}{\left(y+z\right)\left(x+z\right)\left(x+y\right)}=\dfrac{\left(x^3+y^3+z^3+xyz\right)\left(x+y+z\right)}{\left(y+z\right)\left(x+z\right)\left(x+y\right)}=0\)

Answer 2

[Đức Cường]

C2 nè :
\(P=\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}=1\)

\(P=\left(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\right)\left(x+y+z\right)=x+y+z .\)

\(\Leftrightarrow\dfrac{x^2+x\left(y+z\right)}{y+z}+\dfrac{y^2+y\left(x+z\right)}{z+x}+\dfrac{z^2+z\left(x+y\right)}{x+y}=x+y+z.\)

\(\Leftrightarrow\dfrac{x^2}{y+z}+x+\dfrac{y^2}{z+x}+y+\dfrac{z^2}{x+y}+z=x+y+z \left(ĐPCM\right)\)