Does a + B + C greater than or equal to ab + BC + AC have this theorem? Another ··········· a + B + C is greater than AB + BC + AC. is there such a theorem?

Yes! Because: A ^ 2 + B ^ 2 + C ^ 2 = AB + AC + BC, so: 2A ^ 2 + 2B ^ 2 + 2C ^ 2-2ab-2ac-2bc = 0, so: (a-b) ^ 2 + (B-C) ^ 2 + (C-A) ^ 2 = 0, so: A-B = 0, B-C = 0, C-A = 0, so: a = b = C, the second one is not

U = x ^ y + y ^ arctan (x / y) for y partial derivative of the detailed explanation, thank you!

Let P = y ^ arctan (x / y)

On the mathematical derivative g (x) = xlnx-a (x-a), a belongs to R, find the minimum value of the function g (x) in the interval [1, e]
I can find G '(x) and know how to classify it. But if I have a parameter, I can't judge the positive and negative of the derivative. If I have a parameter, how can I judge the positive and negative of the derivative

Because g (x) = xlnx-a (x-a),
So G '(x) = LNX + 1-A
(by drawing its graph, we can see that the function g (x) decreases monotonically. That is to say, taking G '(x) = 0 as the boundary, the function g (x) first increases and then decreases, and there is a maximum value at G' (x) = 0.)
From G '(x) = 0, x = e ^ (1-A)
On the interval [1, e]
1. When e ^ (1-A) = 1, G (x) min = g (E) = e-A (e-A);
2. When 1

After adding the same number to both numerator and denominator, it can be divided into 7 out of 13. What is the added number?

The denominator is 35-17 = 18 larger than the numerator
It's about 7 / 13
The denominator is 13-7 = 6 larger than the numerator
So it's about 18 / 6 = 3
So later the denominator was 13 × 3 = 39
39-35=4
A: the number added is 4

Differential equations of higher numbers
Use appropriate variables to change the following equation into a separable variable equation and find the general solution:
y'=y^2+2(sinx-1)y+(sinx)^2-2sinx-cosx+1

Y '= y ^ 2 + 2 (sinx-1) y + (SiNx) ^ 2-2sinx-cosx + 1 = y ^ 2 + 2 (sinx-1) y + (sinx-1) ^ 2-cosx = (y + sinx-1) ^ 2-cosx, i.e., y' + cosx = (y + sinx-1) ^ 2, let u = y + sinx-1, then the original differential equation is reduced to Du / DX = u ^ 2, the general solution is - 1 / u = x + C, and replace Yu = y + sinx-1

In △ ABC, a = M2 + N2, B = m2-n2, C = 2Mn, and M > n > 0, (1) can you judge the longest side of △ ABC? Please explain the reason; (2) what triangle △ ABC is, please explain it by calculation

(1) A is the longest side because ∵ A-B = (M2 + N2) - (m2-n2) = 2n2 ＞ 0, a-c = (M2 + N2) - 2Mn = (m-n) 2 ＞ 0, ∵ a ＞ B, a ＞ C, ∵ A is the longest side; (2) ABC is a right triangle because

Does a + B + C greater than or equal to ab + BC + AC have this theorem? Another ··········· a + B + C is greater than AB + BC + AC. is there such a theorem?