1, because A > B > 0 and A-B > 0, a^2+ 1/b(a-b)=(a-b+b)^2+ 1/b(a-b)=(a-b)^2+b(a-b)+b(a-b)+(b-a)^2+ 1/4b(a-b)+ 1/4b(a-b)+ 1/4b(a-b)+ 1/4b(a-b) ≥8* 1/ 2 = 4, and the conditions for the equal sign are: a-b=b, b(a-b)= 1/4b(a-b), that is, b 2 =1/2 and a 2 = 2.

2. Because 2 x > 0, let 2 x = t, so f (x) = (2x+4)/(4x+8) = (t+4)/(t2+8) = (t+4)/[(t+4-4). Because B2-3b+21/4 = (b-3/2) 2+3 ≥ 3 >1/(4 √ 6-8) ≥ f (x), the second problem is proved.

3. Because the big side of a triangle corresponds to a big corner, so ∠B is not a big corner, that is, ∠ b is less than 90, so cos ∠ b is greater than 0; (1), the cosine theorem in △ABC is A 2+C 2-2ac * COSB = B 2, that is, 4+C 2-4C * COSB = 1, cosb = (c+3/c) *1. (2) Because COSB = (A2+C2-B2)/2ac ≥ (2ac-B2)/2ac =1/2, b ≤ 60.

4. Use Helen's formula S = √ p (p-a) (p-b) (p-c) = √ 3 * √1/3 * p (p-a) (p-c) ≤ √ 3 * [(1] In the above formula, Helen's formula can be derived from sine and cosine theorems. You can Baidu it. You can also try to prove this problem with trigonometric function, and your result seems to be problematic.

5. Let BF=a, BE=b, easy to know ab= 10, cosB=4/5, and the cosine theorem is: EF 2 = A 2+B 2-2 AB COSB ≥ 2A B-2AB COSB = 20-16 = 4.

Finally finished, so many questions you'd better send to math, if there is a next time!