大家好,乐乐来为大家解答以下的问题，关于三角不等式公式取等条件，三角不等式公式这个很多人还不知道,那么今天让乐乐带着大家一起来看看吧！

竞赛题?1.令A=2x,B=2y,C=2z,则A+B+C=π且A,B,C>0,A,B,C可以看成某个三角形三个顶角，设s=(a+b+c)/2,R是外接圆半径,r是内切圆半径，S是三角形面积8(cosx*cosy*cosz)^2=(1+cosA)(1+cosB)(1+cosC)=(1+(b^2+c^2-a^2)/2bc)(1+(c^2+a^2-b^2)/2ac)(1+(b^2+a^2-c^2)/2ab)=(a+b+c)^3(b+c-a)(c+a-b)(a+c-b)/8a^2b^2c^2=8s^3(s-a)(s-b)(s-c)/a^2b^2c^2=8s^2*S^2/a^2b^2c^2 ...1又因为sin(A/2)sin(B/2)sin(C/2)=r/4R所以27sinxsinysinz=27*r/4R=27*S/4sR ...2由1,2。

即要证:8s^3*4SR≥27a^2b^2c^2因为SR=1/2absinC*c/sinC=abc/4也就是8s^3≥27abc由不等式s=(a+b+c)/2≥3/2(abc)^(1/3)所以s^3≥27/8abc即8s^3≥27abc,得证等号成立当且仅当a=b=c,A=B=C,x=y=z=∏/62.猜想不等式左边≥右边(sinA+sinB+sinC)(1/sinA+1/sinB+1/sinC)=3+(sinB+sinC)/sinA+(sinA+sinC)/sinB+(sinA+sinB)/sinC由sinB+sinC=2sin(B+C)/2cos(B-C)/2≤2sin(B+C)/2=2cosA/2同理sinA+sinC≤2cosB/2,sinA+sinB≤2cosC/2所以(sinA+sinB+sinC)/(1/sinA+1/sinB+1/sinC)≤3+1/sin(A/2)+1/sin(B/2)+1/sin(C/2)只要证:2+2/3(1/sin(A/2)+1/sin(B/2)+1/sin(C/2))≤1/sin(A/2)+1/sin(B/2)+1/sin(C/2)即1/sin(A/2)+1/sin(B/2)+1/sin(C/2)≥6由sin(A/2)+sin(B/2)+sin(C/2)=2sin(A+B)/4cos(A-B)/4+cos(A+B)/2≤2sin(A+B)/4+1-2sin^2(A+B)/4（令sin(A+B)/4=t)=2t+1-2t^2≤3/2再由调和平均≤算术平均:3/(1/sin(A/2)+1/sin(B/2)+1/(sinC/2))≤(sin(A/2)+sin(B/2)+sin(C/2))/3≤1/2即1/sin(A/2)+1/sin(B/2)+1/sin(C/2)≥6,得证等号成立当且仅当A=B=C=∏/3。

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