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Ta có:
\(\left(x+y+z\right)\left(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\right)=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}+x+y+z\)
\(\Leftrightarrow x+y+z=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}+x+y+z\)
\(\Leftrightarrow\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}=0\)
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Câu hỏi của Vũ Anh Quân - Toán lớp 8 | Học trực tuyến nè nhé b .
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=\(Q=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}=\dfrac{x^2\left(z+x\right)\left(x+y\right)+y^2\left(y+z\right)\left(x+y\right)+z^2\left(y+z\right)\left(z+x\right)}{\left(y+z\right)\left(x+z\right)\left(x+y\right)}=\dfrac{\left(x^3+y^3+z^3+xyz\right)\left(x+y+z\right)}{\left(y+z\right)\left(x+z\right)\left(x+y\right)}=0\)
C2 nè :
\(P=\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}=1\)
\(P=\left(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\right)\left(x+y+z\right)=x+y+z .\)
\(\Leftrightarrow\dfrac{x^2+x\left(y+z\right)}{y+z}+\dfrac{y^2+y\left(x+z\right)}{z+x}+\dfrac{z^2+z\left(x+y\right)}{x+y}=x+y+z.\)
\(\Leftrightarrow\dfrac{x^2}{y+z}+x+\dfrac{y^2}{z+x}+y+\dfrac{z^2}{x+y}+z=x+y+z \left(ĐPCM\right)\)
Áp dụng bđt Cauchy, ta có:
\(\dfrac{x^2}{y^2}+\dfrac{y^2}{z^2}+\dfrac{z^2}{x^2}\ge\sqrt{\dfrac{x^2}{y^2}\times\dfrac{y^2}{z^2}}+\sqrt{\dfrac{y^2}{z^2}\times\dfrac{z^2}{x^2}}+\sqrt{\dfrac{x^2}{y^2}\times\dfrac{z^2}{x^2}}=\dfrac{x}{z}+\dfrac{y}{x}+\dfrac{z}{y}\left(\text{đ}pcm\right)\)
Dấu "=" xảy ra khi x = y = z
