解令a/2=b/3=c/4=k
则a=2k,b=3k,c=4k
则(ab+bc+ca)/a2+b2+c2
=(2k*3k+3k*4k+4k*2k)/[(2k)2+(3k)2+(4k)2]
=(6k*k+12k*k+8kk)/[4k2+9k2+16k2]
=26k^2/29k^2
=26/29
解令a/2=b/3=c/4=k
则a=2k,b=3k,c=4k
则(ab+bc+ca)/a2+b2+c2
=(2k*3k+3k*4k+4k*2k)/[(2k)2+(3k)2+(4k)2]
=(6k*k+12k*k+8kk)/[4k2+9k2+16k2]
=26k^2/29k^2
=26/29