Chứng minh nếu \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\) thì (x2+y2+z2).(a2+b2+c2) = (ax+by+cz)2
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đặt x/a=y/b=z/c=k
=>x=a.k,
y=b.k
z=c.k
=>(a^2k^2+b^2k^2+c^2k^2)(a^2+b^2+c^2)=k^2.(a^2+b^2+c^2)^2(1)
(ax+by+cz)^2=(a.a.k+b.b.k+c.c.k)^2=(a^2.k+b^2.k+c^2.k)^2
=k^2(a^2+b^2+c^2)(2)
từ (1)(2)=> nếu x/a=y/b=z/c thì (x2 + y2 + z2) (a2 + b2 + c2) = (ax + by + cz)2
=>
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Có: \(a+b+c=1\Leftrightarrow\left(a+b+c\right)^2=1\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=\dfrac{x+y+z}{a+b+c}\)
\(\Rightarrow\dfrac{x^2}{a^2}=\dfrac{y^2}{b^2}=\dfrac{z^2}{c^2}=\dfrac{\left(x+y+z\right)^2}{\left(a+b+c\right)^2}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}\)
\(\Rightarrow\left(x+y+z\right)^2=x^2+y^2+z^2\) (do \(\left(a+b+c\right)^2=a^2+b^2+c^2=1\))
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Ta có x + y = 2cz + ax + by = 2cz + z
hay 2cz = x + y - z, suy ra c = \(\frac{x+y-z}{2z}\)
do đó: \(1+c=\frac{x+y+z}{2z}\) hay \(\frac{1}{1+c}=\frac{2z}{z+y+z}\)
Tương tự \(1+a=\frac{x+y+z}{2x}\) hay \(\frac{1}{1+a}=\frac{2x}{x+y+z}\)
\(1+b=\frac{x+y+z}{2y}\) hay \(\frac{1}{1+b}=\frac{2y}{x+y+z}\)
Vậy \(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)
Ta có \(\left\{\begin{matrix}x=by+cz\\y=ax+cz\\z=ax+by\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}ax+x=ax+by+cz\\by+y=ax+by+cz\\cz+z=ax+by+cz\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x\left(a+1\right)=ax+by+cz\\y\left(b+1\right)=ax+by+cz\\z\left(c+1\right)=ax+by+cz\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}a+1=\frac{ax+by+cz}{x}\\b+1=\frac{ax+by+cz}{y}\\c+1=\frac{ax+by+cz}{z}\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}\frac{1}{a+1}=\frac{x}{ax+by+cz}\\\frac{1}{b+1}=\frac{y}{ax+by+cz}\\\frac{1}{c+1}=\frac{z}{ax+by+cz}\end{matrix}\right.\)
\(\Rightarrow\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=\frac{x+y+z}{ax+by+cz}\)
Ta lại có \(\left\{\begin{matrix}x=by+cz\\y=ax+cz\\z=ax+by\end{matrix}\right.\)
\(\Rightarrow x+y+z=2\left(ax+by+cz\right)\)
\(\Rightarrow\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=\frac{x+y+z}{ax+by+cz}=\frac{2\left(ax+by+cz\right)}{ax+by+cz}=2\)
Vậy \(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=2\left(đpcm\right)\)
Giả Sử điều ta phải chứng mình là có:
\(\Rightarrow x^2a^2+x^2b^2+x^2c^2+y^2a^2+y^2b^2+y^2c^2+z^2a^2+z^2b^2+z^2c^2=a^2x^2+b^2y^2+c^2z^2+\)
\(2axby+2bycz+2czax\)
\(\Rightarrow a^2x^2-a^2x^2+by^2-b^2y^2+c^2z^2-c^2z^2+x^2b^2+x^2c^2+y^2a^2+y^2c^2+z^2a^2+z^2b^2-\)
\(2axby-2bycz-2czax=0\)
\(\Rightarrow x^2b^2-2axby+a^2y^2+y^2c^2-2bycz+b^2z^2+z^2a^2-2czax+c^2x^2=0\)
\(\Rightarrow\left(xb-ay\right)^2+\left(yc-bz\right)^2+\left(za-cx\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}xb-ay=0\\yc-bz=0\\za-cx=0\end{cases}\Rightarrow}\hept{\begin{cases}xb=ay\\yc=bz\\za=cx\end{cases}\Rightarrow\hept{\begin{cases}\frac{x}{a}=\frac{y}{b}\\\frac{y}{b}=\frac{z}{c}\\\frac{z}{c}=\frac{x}{a}\end{cases}\Rightarrow}\frac{x}{a}=\frac{y}{b}=\frac{z}{c}}\)( mà giả thuyết cho ta x/a=y/b=z/c nên điều ta cần chứng minh đúng)
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