大师作文网三角函数不等式1)已知x+y+z=∏/2,x,y,z>0,求证:8(cosx·cosy·cosz)²≥27si
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三角函数不等式
1)已知x+y+z=∏/2,x,y,z>0,求证:8(cosx·cosy·cosz)²≥27sinx·siny·sinz
2)已知A+B+C=∏,A,B,C>0,试确定1/sin(A/2)+1/sin(B/2)+1/sin(C/2)与(1/sinA+1/sinB+1/sinC)(sinA+sinB+sinC)·2/3的大小关系,并证明.
1)已知x+y+z=π/2,z>0,
求证:8(cosx·cosy·cosz)²≥27sinx·siny·sinz
2)已知A+B+C=π
,A,B,C>0,试确定1/sin(A/2)+1/sin(B/2)+1/sin(C/2)
与(1/sinA+1/sinB+1/sinC)(sinA+sinB+sinC)·2/3的大小关系，
并证明。
1)已知x+y+z=∏/2,x,y,z>0,求证:8(cosx·cosy·cosz)²≥27sinx·siny·sinz
2)已知A+B+C=∏,A,B,C>0,试确定1/sin(A/2)+1/sin(B/2)+1/sin(C/2)与(1/sinA+1/sinB+1/sinC)(sinA+sinB+sinC)·2/3的大小关系,并证明.
1)已知x+y+z=π/2,z>0,
求证:8(cosx·cosy·cosz)²≥27sinx·siny·sinz
2)已知A+B+C=π
,A,B,C>0,试确定1/sin(A/2)+1/sin(B/2)+1/sin(C/2)
与(1/sinA+1/sinB+1/sinC)(sinA+sinB+sinC)·2/3的大小关系，
并证明。
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=∏/6
2.
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
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=∏/6
2.
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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