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Áp dụng BĐT Bunhiacopxki , ta có :
\(\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)\text{≥}\left(ax+by+cz\right)^2\)
\("="\text{⇔}\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\)
⇒ \(\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)\text{=}\left(ax+by+cz\right)^2\)
P/s : Bạn cũng có thể biến đổi VT cũng ra nhé .
2) ta có: \(VT=\left(a^2+b^2\right)\left(x^2+y^2\right)\) và \(VP=\left(ax+by\right)^2\)
tính hiệu của cả VT và VP
suy ra: \(\left(ay+bx\right)^2=0\Rightarrow ay=bx\)
vì \(x,y\ne0\Rightarrow\dfrac{a}{x}=\dfrac{b}{y}\left(đpcm\right)\)
3)(a2+b2+c2)(x2+y2+z2)=(ax+by+cz)2 (1)
biến đổi đẳng thức (1) thành (ay+bx)2 + (bz-cy)2 +(az-cx)2 =0
\(\Rightarrow\) Đpcm
Phương Ann Nhã Doanh Đinh Đức Hùng Mashiro Shiina
Nguyễn Thanh Hằng Nguyễn Huy Tú Lightning Farron
Akai Haruma Võ Đông Anh Tuấn
mấy anh chị cm cho e thêm cái : \(\dfrac{ay+bx}{c}=\dfrac{bz+cy}{a}=\dfrac{cx+az}{b}\)
\(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)=\left(ax+by+cz\right)^2\)
=>\(a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2=a^2x^2+b^2y^2+c^2z^2+2axby+2bycz+2czax\)
=>\(a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2=2axby+2bycz+2czax\)
=>\(\left(a^2y^2-2aybx+b^2x^2\right)+\left(b^2z^2-2bzcy+c^2y^2\right)+\left(c^2x^2-2cxaz+a^2z^2\right)=0\)
=>\(\left(ay-bx\right)^2+\left(bz-cy\right)^2+\left(cx-az\right)^2=0\)
Mà \(\hept{\begin{cases}\left(ay-bx\right)^2\ge0\\\left(bz-cy\right)^2\ge0\\\left(cx-az\right)^2\ge0\end{cases}\Rightarrow\left(ay-bx\right)^2+\left(bz-cy\right)^2+\left(cx-az\right)^2\ge0}\)
\(\Rightarrow\hept{\begin{cases}ay-bx=0\\bz-cy=0\\cx-az=0\end{cases}}\Rightarrow\hept{\begin{cases}ay=bx\\bz=cy\\cx=az\end{cases}\Rightarrow\hept{\begin{cases}\frac{a}{x}=\frac{b}{y}\\\frac{b}{y}=\frac{c}{z}\\\frac{c}{z}=\frac{a}{x}\end{cases}\Rightarrow}\frac{a}{x}=\frac{b}{y}=\frac{c}{z}}\) (đpcm)
\(\dfrac{\left(ax+by+cz\right)^2}{x^2+y^2+x^2}=a^2+b^2+c^2\)
\(\Leftrightarrow\left(x^2+y^2+x^2\right)\left(a^2+b^2+c^2\right)=\left(ax+by+cz\right)^2\)\(\Leftrightarrow a^2x^2+b^2x^2+c^2x^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+x^2z^2=a^2x^2+b^2y^2+c^2z^2+2axby+2axcz+2bycz\)\(\Leftrightarrow\left(a^2y^2+2axby+b^2x^2\right)+\left(a^2z^2+2axcz+c^2x^2\right)+\left(b^2z^2+2bycz+c^2y^2\right)=0\)\(\Leftrightarrow\left(ay+bx\right)^2+\left(az+cx\right)^2+\left(bz+cy\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}ay=bx\\az=cx\\bz=cy\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}\dfrac{a}{x}=\dfrac{b}{y}\\\dfrac{a}{x}=\dfrac{c}{z}\\\dfrac{b}{y}=\dfrac{c}{z}\end{matrix}\right.\Leftrightarrow\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\left(đpcm\right)\)
\(1.\) Giả sử : \(a\ge b\ge c\Rightarrow a+b\ge a+c\ge b+c\)
Ta có : \(\dfrac{c}{a+b}\le\dfrac{c}{b+c};\dfrac{b}{a+c}\le\dfrac{b}{b+c};\dfrac{a}{b+c}=\dfrac{a}{b+c}\)
\(\Rightarrow\dfrac{c}{a+b}+\dfrac{b}{a+c}+\dfrac{a}{b+c}\le\dfrac{b+c}{b+c}+\dfrac{a}{b+c}=1+\dfrac{a}{b+c}< 1+1=2\left(đpcm\right)\)
\(2.\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\)
\(\Leftrightarrow\dfrac{yz+xz+xy}{xyz}=\dfrac{1}{x+y+z}\)
\(\Leftrightarrow\left(x+y+z\right)\left(xy+yz+xz\right)=xyz\)
\(\Leftrightarrow x^2y+x^2z+xy^2+y^2z+xyz+xyz+yz^2+xz^2=0\)
\(\Leftrightarrow xy\left(x+y+z\right)+yz\left(x+y+z\right)+xz\left(x+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)y\left(x+z\right)+xz\left(x+z\right)=0\)
\(\Leftrightarrow\left(x+z\right)\left(xy+y^2+yz+xz\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(x+z\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-y\\y=-z\\x=-z\end{matrix}\right.\)
+) Với : \(x=-y\) , ta có :
Đpcm \(\Leftrightarrow-\dfrac{1}{y^{2011}}+\dfrac{1}{y^{2011}}+\dfrac{1}{z^{2011}}=\dfrac{1}{-y^{2011}+y^{2011}+z^{2011}}\)
\(\Leftrightarrow\dfrac{1}{z^{2011}}=\dfrac{1}{z^{2011}}\left(luôn-đúng\right)\)
Tương tự với 2 TH còn lại .
\(\RightarrowĐCPM\)
Đặt \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=k\Rightarrow x=ak;y=bk;z=ck\)
Ta có:
\(\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)=\left[\left(ak\right)^2+\left(bk\right)^2+\left(ck\right)^2\right]\left(a^2+b^2+c^2\right)=k^2\left(a^2+b^2+c^2\right)^2\)
\(\left(ax+by+cx\right)^2=\left(a^2k+b^2k+c^2k\right)^2=k^2\left(a^2+b^2+c^2\right)^2\)
=> đpcm
\(\left\{{}\begin{matrix}\dfrac{x}{a}=\dfrac{y}{b}\\\dfrac{x}{a}=\dfrac{z}{c}\\\dfrac{y}{b}=\dfrac{z}{c}\end{matrix}\right.\) \(\Rightarrow\) \(\left\{{}\begin{matrix}ay=bx\\az=cx\\bz=cy\end{matrix}\right.\)
\(\Rightarrow\) \(\left\{{}\begin{matrix}\left(ay-bx\right)^2=0\\\left(az-cx\right)^2=0\\\left(bz-cy\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\) \(\left(ay-bx\right)^2+\left(az-cx\right)^2+\left(bz-cy\right)^2=0\)
\(a^2y^2-2axby+b^2x^2+a^2z^2-2axcz+c^2x^2+b^2z^2-2bycz+c^2y^2=0\)
\(a^2y^2+b^2x^2+a^2z^2+c^2x^2+b^2z^2+c^2y^2-\left(2axby+2bycz+2axcz\right)=0\)
\(a^2y^2+b^2x^2+a^2z^2+c^2x^2+b^2z^2+c^2y^2+a^2x^2+b^2y^2+c^2z^2-\left(a^2x^2+b^2y^2+c^2z^22axby+2bycz+2axcz\right)=0\)
\(\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)-\left(ax+by+cz\right)^2=0\)
\(\Rightarrow\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)=\left(ax+by+cz\right)^2\) ( đpcm )
Mệt chết :VV