已知x,y,z属于(0,派/2),sin^2x+sin^2y+sin^2z=1,求(sinx+siny+sinz)/(c

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  • x,y,z属于(0,派/2)

    sinx,cosx∈(0,1)

    对于a>0,b>0,有不等式:开根号下(a^2+b^2)≥根号2*(a+b)/2

    sin^2x+sin^2y+sin^2z=1

    cosx=开根号下(sin^2y+sin^2z)≥根号2*(siny+sinz)/2

    cosy=开根号下(sin^2x+sin^2z)≥根号2*(sinx+sinz)/2

    cosz=开根号下(sin^2x+sin^2y)≥根号2*(sinx+siny)/2

    仅当sinx=siny=sinz的时候,三式的等号成立.

    三式相加得,cosx+cosy+cosz≥根号2*(sinx+siny+sinz)

    所以(sinx+siny+sinz)/(cosx+cosy+cosz)≤根号2/2

    仅当sinx=siny=sinz=根号3/3时,(sinx+siny+sinz)/(cosx+cosy+cosz)取最大值根号2/2