在数列中请用数学归纳法证明;1/（A1+B2)+1/(A2+B2)····+1/(AN+BN)<5/12

（A1+B1)+1/(A2+B2)••••+1/(AN+BN)+1/(A(N+1)+B(N+1))
楼上的,你的家长很强大,这些都懂

为了利用归纳法证明，需将不等式改为更强的形式
1/(A1+B1)+1/(A2+B2)…+1/(AN+BN)≤5/12-1/(2(N+1))
使不等式右边也含有N，显然上式成立，原不等式也成立。
当N=1时,1/(A1+B1)=1/(2+4)=1/6≤5/12-1/4,不等式成立。
假设对任意N不等式成立,考虑N+1时的情况，由归纳法假设得
1/（A1+B1)+1/(A2+B2)••••+1/(AN+BN)+1/(A(N+1)+B(N+1))
< 5/12-1/(2(N+1))+ 1/(A(N+1)+B(N+1))
= 5/12-1/(2(N+1))+1/((N+1)(N+2)+(N+2)^2)
由(N+1)(N+2)<(N+2)^2, 得
2(N+1)(N+2)<(N+1)(N+2)+(N+2)^2
-1/(2(N+1)(N+2))<-1((N+1)(N+2)+(N+2)^2),
1/(2(N+2))-1/(2(N+1))<-1((N+1)(N+2)+(N+2)^2),
-1/(2(N+1))+1/((N+1)(N+2)+(N+2)^2)<-1/(2(N+2)),
5/12-1/(2(N+1))+1/((N+1)(N+2)+(N+2)^2)<5/12-1/(2(N+2)), 故得
1/（A1+B1)+1/(A2+B2)…+1/(AN+BN)+1/(A(N+1)+B(N+1))<5/12-1/(2(N+2))
完成归纳法证明。

可以问家长/（A1+B1)+1/(A2+B2)••••+1/(AN+BN)+1/(A(N+1)+B(N+1))

对呀