已知b1/a1>d1/c1,b2/a2>d2/c2,求证(b1+b2)/(a1+a2)>(d1+d2)/(c1+c2)
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已知b1/a1>d1/c1,b2/a2>d2/c2,求证(b1+b2)/(a1+a2)>(d1+d2)/(c1+c2)
![已知b1/a1>d1/c1,b2/a2>d2/c2,求证(b1+b2)/(a1+a2)>(d1+d2)/(c1+c2)](/uploads/image/z/5681227-67-7.jpg?t=%E5%B7%B2%E7%9F%A5b1%2Fa1%3Ed1%2Fc1%2Cb2%2Fa2%3Ed2%2Fc2%2C%E6%B1%82%E8%AF%81%28b1%2Bb2%29%2F%28a1%2Ba2%29%3E%28d1%2Bd2%29%2F%28c1%2Bc2%29)
设e1,e2,分别与c1,c2同号,且满足
b1/a1=(d1+e1)/c1,b2/a2=(d2+e2)/c2
则由等比关系可得
(b1+b2)/(a1+a2)=(d1+d2+e1+e2)/(c1+c2)
=(d1+d2)/(c1+c2)+(e1+e2)/(c1+c2)
>(d1+d2)/(c1+c2)
b1/a1=(d1+e1)/c1,b2/a2=(d2+e2)/c2
则由等比关系可得
(b1+b2)/(a1+a2)=(d1+d2+e1+e2)/(c1+c2)
=(d1+d2)/(c1+c2)+(e1+e2)/(c1+c2)
>(d1+d2)/(c1+c2)
已知b1/a1>d1/c1,b2/a2>d2/c2,求证(b1+b2)/(a1+a2)>(d1+d2)/(c1+c2)
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