恒等式证明已知：a,b,c为三角形ABC三边,且2a^2/(1+a^2)=b,2b^2/(1+b^2)=c,2c^2/(

全部取倒数得,
1/b=1/2a^2+1/2,
1/c=1/2b^2+1/2,
1/a=1/2c^2+1/2,
三试相加,配方得,
1/2(1/a^2-2/a+1)+1/2(1/b^2-2/b+1)+1/2(1/c^2-2/c+1)=0,
即1/2(1/a-1)^2+1/2(1/b-1)^2+1/2(1/c-1)^2=0,
所以1/a-1=1/b-1=1/c-1.
即:a=b=c=1,
所以△ABC是等边三角形,