Lời giải:
Cộng 3 PT lại ta có:
$x(a+b+c)+y(a+b+c)=a+b+c$
$\Leftrightarrow (a+b+c)(x+y-1)=0$
$\Rightarrow a+b+c=0$ hoặc $x+y-1=0$
TH1: $a+b+c=0\Leftrightarrow a+b=-c$
Khi đó: $a^3+b^3+c^3=(a+b)^3-3ab(a+b)+c^3$
$=(-c)^3-3ab(-c)+c^3=3abc$
$\Rightarrow \frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=3$ (đpcm)
TH2: $x+y-1=0\Leftrightarrow y=1-x$
Thay vô hpt \(\left\{\begin{matrix} ax+b(1-x)=c\\ bx+c(1-x)=a\\ cx+a(1-x)=b\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x(a-b)=c-b\\ x(b-c)=a-c\\ x(c-a)=b-a\end{matrix}\right.\)
\(\Rightarrow x^3(a-b)(b-c)(c-a)=(c-b)(a-c)(b-a)=-(a-b)(b-c)(c-a)\)
\(\Leftrightarrow (a-b)(b-c)(c-a)(x^3+1)=0\)
Nếu $a-b=0$ thì kéo theo $b-c=c-a=0$
$\Rightarrow a=b=c$
Nếu $b-c=0; c-a=0$ thì tương tự
Nếu $x^3+1=0\Leftrightarrow x=-1$
$\Rightarrow b-a=c-b=a-c\Rightarrow a=b=c$
Tóm lại $a=b=c$
Do đó: $\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=1+1+1=3$ (đpcm)
