Solution: The problem of 1 needs to be seen clearly. The reduction to absurdity proves that ab is under the root sign of half a+b≥, not that A is greater than 0 and B is greater than 0, but that AB is included in it. There is only one condition that makes √ab true. You can change it to a condition that makes √ab hold. What we want to prove is ab under the radical sign of semi-a+b≥

Suppose a+b/2

Then a+b

(√a)^2-2√ab+(√b)^2<； 0(√a-√b)^2<； 0

Because the square of any number is not negative, the assumption is not valid.

Then a+b/2 >; =√ab

2. Here I supplement it with the second person's answer.

Proof: suppose there is an integer m, and n makes m? =n？ +1998, so there are four cases of m and n.

(1) m and n are odd numbers, so m = 2k+ 1)m, n = 2t+ 1 (k, t ∈ z) (2k+ 1)? =(2t+ 1)？ + 1998

Simplify: (k? +k)-(t？ +t)=499.5 and k? +k and t? +t is an integer, integer minus integer = integer, and the original formula = decimal, so the hypothesis cannot be refuted.

(2) m and n are even numbers, so m = 2)m, n = 2tk, t ∈ z4k? =4t？ +1998 proves that 4k likes it (1)? -4t？ As an integer, it contradicts the assumption = 1998/4, so it is impossible to contradict the assumption.

(3)m is odd and n is even, so m=2k+ 1 n=2t k, t∈z 4(k? +k)=4t？ +1997 In the same way, it can be concluded that contradiction and hypothetical contradiction are impossible.

(4)m is even and n is odd, so m=2k n=2t k, t∈z 4k? =4(t？ +t)+ 1999 The same reason does not conform to the hypothesis.

To sum up, the hypothesis is not established, and the original proposition is established.

(3) With regard to this condition, comparison within the scope of inequality is generally found by subsets, which is proved as follows:

Q proposition simplification (the process of simplification is equivalent deformation)

[x-(1+m)] [x-(1-m)] ≤ 0 (m is greater than 0)

This inequality has a solution because it can decompose the factor:

Because m is greater than 0,1-m <1+m.

Finally, it is 1-m≤x≤ 1+m (at this time, as long as x satisfies this inequality, the original inequality holds).

Q is the necessary and sufficient condition of p.

This uses subsets:

Let a = {x |-5 ≤ x ≤ 7} b = {x |1-m ≤ x ≤1+m}

So B is really included in A (that is, the value satisfying B must satisfy A, but the value satisfying A does not necessarily satisfy B, which is equivalent to the necessary and sufficient condition that Q is P).

So1-m >; -5 1+m & lt； 7 so m

M & gt0

So to sum up, 0 < m < 6.

I have done this problem, but I don't think the method is suitable for most problems, but this problem is a bit unconventional.

It is proved that if the roots of quadratic equations ① and ② are integers, then these two equations must have roots:

(-4)^2-4m*4>； =0 and (4m) 2-4 * (4m 2-4m-5) > =0, and m is not equal to 0.

The solution is-5/4 < = m <; = 1 and m is not equal to 0.

M is an integer, so m=- 1 or 1.

Enter the equation test respectively:

De: m= 1

It's over! ! ! ! ! !