Prove the inequality for any real number A.B A2 + B2 + 4 \ (A2 + B2 + 1) ≥ 3 and explain under what conditions to take the equal sign

Prove the inequality for any real number A.B A2 + B2 + 4 \ (A2 + B2 + 1) ≥ 3 and explain under what conditions to take the equal sign

Yes a ²+ b ²+ [4/(a ²+ b ²+ 1) ] ≥ 3!
a ²+ b ²+ [4/(a ²+ b ²+ 1)]
=a ²+ b ²+ 1+[4/(a ²+ b ²+ 1)]-1
≥2√{(a ²+ b ²+ 1)·[4/(a ²+ b ²+ 1)]} -1=3
If and only if a ²+ b ²+ 1=4/(a ²+ b ²+ 1) Namely a ²+ b ²= The 1:00 sign holds

Let a, B and C be positive numbers, and prove: BC a+ca b+ab c≥a+b+c．

Certificate: ∵ 2 (BC)
a+ac
b+ab
c）
=（bc
a+ac
b）+（bc
a+ab
c）+（ac
b+ab
c）
≥2
abc2
ab+2
acb2
ac+2
bca2
bc
=2c+2b+2a，
∴bc
a+ac
b+ab
c≥a+b+c
If and only if a = b = C, the equal sign holds

a. B is a positive number, prove that the root sign AB is greater than or equal to 2 / (1 / A + 1 / b) (proved by basic inequality)

a+b>=2√(ab)
1 / (a + b) 0, b > 0 (multiply both sides by 2Ab)
2ab/(a+b)

Let ab ≠ 0 and prove (B / a) + (A / b) = (b ^ 2 + A ^ 2) / AB ≥ 2Ab / AB = 2 by using the basic inequality. Point out the error of this proof and correct it

(a-b) ² ≥0
a ²+ b ²- 2ab≥0
a ²+ b ² ≥2ab
(a ²+ b ²)/ ab≥2 ab≠0
a/b+b/a≥2
(b/a)+(a/b)=(b^2+a^2)/ab≥2

Prove the inequality | a + B | / 1 + | a + B|

Proof: [method 1] 1. Let | a + B ≠ 0
Then, | a | + | B | | a + B | > 0
Therefore, 1 / (|a| + |b|) 1 / | a + B|
Therefore, 1 / (|a| + |b| + 1 "1 / | a + b| + 1"
Therefore, (1 + | a | + | B |) / (| a | + | B |) (1 + | a + B |) / | a + B|
Therefore, the above formula is reversed: (|a| + |b|) / (1 + |a| + |b|) |a + b| / (1 + | a + b|)
Namely: | a + b| / 1 + | a + B|

What is the derivative of y = (2x square + 3) (3x-2)?

y=(2x^2+3)(3x-2)
Derivation:
=4x(3x-2)+3(2x^2+3)
=12x^2-8x+6x^2+9
=18x^2-8x+9