giải:
ta có : x/a+y/b+z/c =1
<=> (x/a+y/b+z/c )^2=1
<=>x^2/a^2 +y^2/b^2+z^2 /c^2 +2( xy/ab +yz/bc +xz/ac)=1 (1)
ta lại có :a/x+b/y+c/z=0
<=>ayz+bxz+cxy=0
<=>xy/ab +yz/bc +xz/ac=0 (2)
thay (2) vào (1) ta có :
x^2/a^2 +y^2/b^2+z^2 /c^2 +2( xy/ab +yz/bc +xz/ac)=1
<=> x^2/a^2 +y^2/b^2+z^2 /c^2+2.0=1
<=> x^2/a^2 +y^2/b^2+z^2 /c^2=1 (đpcm)