Question

x,y,z分别为三角形的三边 求证(x2+y2-z2)/2xy+(x2+z2-y2)/2xz+(y2+z2-x2)/2y

Answer 1

(x2+y2-z2)/2xy+(x2+z2-y2)/2xz+(y2+z2-x2)/2yz>1等价于

COSA+COSB+COSC>1等价于COSA+COSB>1-COSC等价于2COS(A+B)/2 *cos(A-B)/2>2(sin(C/2))^2因为A+B)/2+C/2=90度 即证2COS(A+B)/2 *cos(A-B)/2>2(cos(A+B/2))^2

即证cos(A-B)/2>cos(A+B/2

即证cosA/2*cosB/2+sinA/2*sinB/2>cosA/2*cosB/2-sinA/2*sinB/2

即证2sinA/2*sinB/2>0

因为A/2,B/2都是锐角,显然成立