Nếu cái j?

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

=>

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 \(\left(a+b+c\right)^2=a^2+b^2+c^2=1\))

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 \(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=2\left(đpcm\right)\)