已知x,y,z属于(0,派/2),sin^2x+sin^2y+sin^2z=1,求(sinx+siny+sinz)/(cosx+cosy+cosz)的最大值

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