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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
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Lời giải:
Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=m\Rightarrow x=am; y=bm; z=cm\)
Khi đó:
\((x^2+2y^2+3z^2)(a^2+2b^2+3c^2)=[(am)^2+2(bm)^2+3(cm)^2](a^2+2b^2+3c^2)\)
\(=m^2(a^2+2b^2+3c^2)^2(1)\)
Và:
\((ax+2by+3cz)^2=(a.am+2b.bm+3c.cm)^2=[m(a^2+2b^2+3c^2)]^2\)
\(=m^2(a^2+2b^2+3c^2)^2(2)\)
Từ (1) và (2) ta có đpcm.
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Đề sai rồi bạn
Vế phải : \(\left(ax+aby+3cz\right)^2\)
\(=a^2x^2+a^2b^2y^2+9c^2z^2+2a^2bxy+6abcyz+6axcz\)
Vế trái : \(\left(x^2+2y^2+3z^2\right)\left(a^2+2b^2+3c^2\right)\)
\(=\left(x^2+2y^2+3z^2\right)a^2+\left(x^2+2y^2+3z^2\right)2b^2+\left(x^2+2y^2+3z^2\right)3c^2\)
\(=x^2a^2+2a^2y^2+3a^2z^2+2b^2x^2+4b^2y^2+6b^2z^2+3x^2c^2+6c^2y^2+9c^2z^2\)
Ko hề bằng nhau
\(\Rightarrow\)đề sai
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\(\left(2\right)\Rightarrow dpcm\)