cmr a/x=b/y=c/z=a+2b-3c/x+2y+3z
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\(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}=\frac{2b}{2y}=\frac{-3c}{-3z}=\frac{a+2b-3c}{x+2y-3z}\)
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Cần cm: \(\left(x^2+2y^2+3z^2\right)\left(a^2+2b^2+3c^2\right)=\left(ax+2by+3cz\right)^2\)
Theo bđt Cauchy-Schwarz:
\(VT=\left(x^2+2y^2+3z^2\right)\left(a^2+2b^2+3c^2\right)\ge\left(ax+\sqrt{2}y.\sqrt{2}b+\sqrt{3}z.\sqrt{3}c\right)^2\)
\(\Rightarrow VT\ge\left(ax+2by+3cz\right)^2\)\(=VP\)
Dấu "=" khi \(\frac{x}{a}=\frac{\sqrt{2}y}{\sqrt{2}b}=\frac{\sqrt{3}z}{\sqrt{3}c}\Leftrightarrow\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)
Ta thấy dấu "=" ở đây xảy ra vì từ gt \(\Rightarrow\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)
\(\Rightarrowđpcm\)
Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\)
\(\Rightarrow x=ak;y=bk;z=ck\)
\(\left(x^2+2y^2+3z^2\right)\left(a^2+2b^2+3c^2\right)\)
\(=\left[\left(ak\right)^2+2\left(bk\right)^2+3\left(ck\right)^2\right]\left(a^2+2b^2+3c^2\right)\)
\(=k^2\left(a^2+2b^2+3c^2\right)\left(a^2+2b^2+3c^2\right)\)
\(=k^2\left(a^2+2b^2+3c^2\right)^2\left(1\right)\)
\(\left(ax+2by+3cz\right)^2\)
\(=\left(a.ak+2b.bk+3c.ck\right)^2\)
\(=\left[k\left(a^2+2b^2+3c^2\right)\right]^2\)
\(=k^2\left(a^2+2b^2+3c^2\right)^2\left(2\right)\)
Từ \(\left(1\right)\)và\(\left(2\right)\Rightarrow dpcm\)
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Vì \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)
\(\Rightarrow\frac{x}{a}=\frac{4x}{4a}=\frac{2y}{2b}=\frac{5y}{5b}=\frac{3z}{3c}=\frac{6z}{6c}\)
Áp dụng tc của dãy tỉ số bằng nhau ta có :
\(\frac{x}{a}=\frac{4x}{4a}=\frac{2y}{2b}=\frac{5y}{5b}=\frac{3z}{3c}=\frac{6z}{6c}=\frac{x+2y-3z}{a+2b-3c}=\frac{4x-5y+6z}{4a-5b+6c}\)
\(\Rightarrow\frac{x+2y-3z}{4x-5y+6z}=\frac{a+2b-3c}{4a-5b+6c}\left(đpcm\right)\)
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#)Giải :
Đặt \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}=k\Rightarrow\hept{\begin{cases}a=kx\\b=ky\\c=kz\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\left(a^2+2b^2+3c^2\right)\left(x^2+2y^2+3z^2\right)=\left[\left(kx\right)^2+2\left(ky\right)^2+3\left(kz\right)^2\right]\left(x^2+2y^2+3z^2\right)=k^2\left(a^2+2b^2+3c^2\right)^2\left(1\right)\\\left(ax+2by+3cz\right)^2=\left(kx.x+2ky.y+3kz.z\right)^2=\left[k\left(a^2+2b^2+3c^2\right)\right]^2=k^2\left(a^2+2b^2+3c^2\right)^2\left(2\right)\end{cases}}\)
Từ (1) và (2) => đpcm