1、 ABC are all positive numbers, and a + B + C = 1 1/(a+b)+1/(b+c)+1/(a+c)>=9/2 2、 ABC is a positive number Prove a ^ 2 / B + B ^ 2 / C + C ^ 2 / a > = a + B + C

1、 ABC are all positive numbers, and a + B + C = 1 1/(a+b)+1/(b+c)+1/(a+c)>=9/2 2、 ABC is a positive number Prove a ^ 2 / B + B ^ 2 / C + C ^ 2 / a > = a + B + C

The first direct Cauchy is OK, 2 (a + B + C) * [1 / (a + b) + 1 / (B + C) + 1 / (a + C)] > = (1 + 1 + 1) ^ 2, so 1 / (a + b) + 1 / (B + C) + 1 / (a + C) > = 9 / 2 can also be replaced by 1, and then the mean inequality is used. The second mean or ranking can be a ^ 2 / B + b > = 2Ab ^ 2 / C + C > = 2BC ^ 2 / A + a > = 2c3

1. Given a > 0, b > 0 and a + B = 1, then the minimum value of ((1 / A ^ 2) - 1) ((1 / b ^ 2) - 1) is——
2. If a > b > 0, M = (√ a) - (√ b), n = √ (a-b), then the relationship between M and N is——
3. Let a > b > C > 0, P = √ ((a + C) ^ 2 + B ^ 2), q = √ (a ^ 2 + (B + C) ^ 2), s = √ ((a + C) ^ 2)
+b) 2 + C ^ 2), then the smallest of P, Q and S is——
4. Given that C > 1, if M = √ (c + 1) - √ C n = √ C - √ (C-1), then the relationship between M and N is——

2. Let a = 9, B = 4 satisfy the condition, then M = 1, n = √ 5 = 2.236
3. If a = 3, B = 2, C = 1, then p = √ 20, q = √ 18, s = √ 26, the smallest is Q
Open another square, P = √ (a ^ 2 + B ^ 2 + C ^ 2 + 2Ac)
q=√(a^2+b^2+c^2+2bc)
s=√(a^2+b^2+c^2+2ab)
The open root sign does not affect the size of the number; therefore, except for the same part, we will compare which of 2Ac, 2BC, 2Ab is the smallest; since a > b > C > 0, obviously 2BC is the smallest, so q is the smallest
4. Let C be equal to any number greater than 1, for example, let C = 2
1. The minimum value is 9. It's troublesome to prove it. I'm still thinking about it

Ask a mathematical problem of inequality in Senior Two
X < 4 / 5, y = 4x-2 + 1 / 4x-5?

It should be x < 5 / 4,
4x-50
y=4x-2+1/4x-5
=(4x-5)+1/(4x-5)+3
=-[(5-4x)+1/(5-4x)]+3

A mathematical problem about inequality in senior two,
Given that g = lg9 * lg11, compare the size of G and 1

There may be some mistakes in the answers
The second floor should be 0