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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
\(VT=\dfrac{x^2}{x^2+2xy+3zx}+\dfrac{y^2}{y^2+2yz+3xy}+\dfrac{z^2}{z^2+2zx+3yz}\)
\(VT\ge\dfrac{\left(x+y+z\right)^2}{x^2+y^2+z^2+5xy+5yz+5zx}=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+3\left(xy+yz+zx\right)}\ge\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+\left(x+y+z\right)^2}=\dfrac{1}{2}\)
1.
Ta có:
\(x^4+y^4\ge\dfrac{1}{2}\left(x^2+y^2\right)^2=\dfrac{1}{2}\left(x^2+y^2\right)\left(x^2+y^2\right)\ge\left(x^2+y^2\right)xy\)
Đặt vế trái của BĐT cần chứng minh là P, áp dụng bồ đề vừa chứng minh ta có:
\(P\le\dfrac{a.abc}{bc\left(b^2+c^2\right)+a.abc}+\dfrac{b.abc}{ca\left(c^2+a^2\right)+b.abc}+\dfrac{c.abc}{ab\left(a^2+b^2\right)+c.abc}\)
\(P\le\dfrac{a^2.bc}{bc\left(a^2+b^2+c^2\right)}+\dfrac{b^2.ac}{ca\left(a^2+b^2+c^2\right)}+\dfrac{c^2.ab}{ab\left(a^2+b^2+c^2\right)}=1\)
Dấu "=" xảy ra khi \(a=b=c=1\)
2.
\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}=1\)
Dấu "=" xảy ra khi \(x=y=z=\dfrac{2}{3}\)
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 :
\(P=\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}=1\)
\(P=\left(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\right)\left(x+y+z\right)=x+y+z .\)
\(\Leftrightarrow\dfrac{x^2+x\left(y+z\right)}{y+z}+\dfrac{y^2+y\left(x+z\right)}{z+x}+\dfrac{z^2+z\left(x+y\right)}{x+y}=x+y+z.\)
\(\Leftrightarrow\dfrac{x^2}{y+z}+x+\dfrac{y^2}{z+x}+y+\dfrac{z^2}{x+y}+z=x+y+z \left(ĐPCM\right)\)
