đặt a/2014=b/2015=c/2016=k
=>a=2014k;b=2015k;c=2016k
=>4(a-b)(b-c)=4(2014k-2015k)(2015k-2016k)
=4.k(2014-2015).k92015-2016)=4.k.(-1).k.(-1)=4.k^2(1)
=>(c-a)(c-a)=(c-a)^2=(2016k-2014k)(2016k-2014k)=[k(2016-2014)]^2=[k.2]^2=k^2.4(2)
từ (1)và (2)=>4(a-b)(b-c) = (c-a).(c-a)