若x,y,z∈R,且x+y+z=xyz,
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若x,y,z∈R,且x+y+z=xyz,
求证:(y+z)/x+(z+x)/y+(x+y)/z≥2(1/x+1/y+1/z)^2
求证:(y+z)/x+(z+x)/y+(x+y)/z≥2(1/x+1/y+1/z)^2
设
1/x=p
1/y=q
1/z=r
则pq+qr+pr=1
(y+x)/z+(y+z)/x+(z+x)/y≥2(1/x+1/y+1/z)^2
为(pq+qr+pr)[r/p+r/q+q/r+q/p+p/r+p/q]>=2(p+q+r)^2
即2(r^2+p^2+q^2+pq+qr+rp)+rrq/p+rrp/q+qqr/p+qqp/r+ppr/q+ppq/r>=2(p+q+r)^2
即rrq/p+rrp/q+qqr/p+qqp/r+ppr/q+ppq/r>=2(pq+qr+pr)
又因为rrp/q+rrq/p>=2rr
所以只需证rr+pp+qq>=pq+qr+pr
(r-p)^2+(p-q)^2+(q-r)^2>=0
1/x=p
1/y=q
1/z=r
则pq+qr+pr=1
(y+x)/z+(y+z)/x+(z+x)/y≥2(1/x+1/y+1/z)^2
为(pq+qr+pr)[r/p+r/q+q/r+q/p+p/r+p/q]>=2(p+q+r)^2
即2(r^2+p^2+q^2+pq+qr+rp)+rrq/p+rrp/q+qqr/p+qqp/r+ppr/q+ppq/r>=2(p+q+r)^2
即rrq/p+rrp/q+qqr/p+qqp/r+ppr/q+ppq/r>=2(pq+qr+pr)
又因为rrp/q+rrq/p>=2rr
所以只需证rr+pp+qq>=pq+qr+pr
(r-p)^2+(p-q)^2+(q-r)^2>=0
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