Given that a + b > 0, prove that a ^ 3-A ^ 2b is greater than AB ^ 2-B ^ 3

The square of A-B needs to discuss that a is not equal to B?? Otherwise, it should be greater than or equal to

Let a > b > 0, then the minimum value of A2 + 1ab + 1a (a-b) is ()
A. 1B. 2C. 3D. 4

A2 + 1ab + 1a (a-b) = AB + 1ab + a (a-b) + 1a (a-b) ≥ 4 if and only if AB = 1aba (a-b) = 1a (a-b) takes the equal sign, that is, a = 2B = 22 takes the equal sign

Given a > b > 0, find the minimum value of a ^ 2 + 16 / (ab-b ^ 2)

The story of Gauss

1. Gauss is a famous German scientist. His most famous story is that when he was 10 years old, his primary school teacher had an arithmetic problem: calculate 1 + 2 + 3 + + 100 =? It's very difficult for the children who have just learned mathematics. They are adding up the numbers one by one according to the requirements of the questions. But at this time, there is a high

The story of Gauss
The story mainly tells that Gauss's teacher gave him a very difficult problem, and then Gauss seemed to spend half a day or a few days to solve the problem. The teacher asked Gauss, "yes, I'm stupid. It took me so long to solve it." the teacher said, "Newton didn't solve it, Archimedes didn't solve it."
It's not a loss if it's cancelled

One day in 1796, at the University of Gottingen, Germany, a 19-year-old with great mathematical talent finished dinner and began to do three routine mathematical problems assigned to him by his tutor. The first two problems were successfully completed in two hours. The third one was written on another note: only a compass and a straight bar without scale were required

The story of mathematician Gauss

Gauss, an eight year old German scientist, discovered the mathematical theorem. He was born in a poor family. He learned to calculate by himself before he could speak. One night when he was three years old, he watched his father correct his father's mistakes when he was calculating his wages. When he grew up, he became the most outstanding astronomer of the time

Reading material, mathematician Gauss once studied such a problem in school, 1 + 2 + 3 +. N =?) 343400

Is it the sum of 1 + to n? If it is, then I remember correctly that Gauss did it in reverse order
sn=1+2+3+.+(n-2)+(n-1)+n
sn=n+(n-1)+(n-2)+.+3+2+1
Add to get 2Sn = n (n + 1)
So Sn = n (n + 1) divided by 2

Reading materials, the great mathematician Gauss once studied such a problem in school: 1 + 2 + 3 + +100=?
After research, the general conclusion of this problem is 1 + 2 + 3 + +Now let's study a similar problem: 1 × 2 + 2 × 3 + a（a+1）=?
Look at three special equations:
1 × 2 = 1 / 3 (1 × 2 × 3-0 × 1 × 2)
2 × 3 = 1 / 3 (2 × 3 × 4-1 × 2 × 3)
3 × 4 = 1 / 3 (3 × 4 × 5-2 × 3 × 4)
By adding the two sides of the equation, we can get 1 × 2 + 2 × 3 + 3 × 4 = 1 × 3 × 4 × 5 = 20
After reading this material,
（1）1×2+2×3+.+19×20=_
（2）1×2+2×3+.+a·(a+1)=_
（3）1×2×3+2×3×4+...+（a-1）·a·（a+1）=___

In order to type fast, * represents (1) 1 × 2 + 2 × 3 +. + 19 × 20 = 1 / 3 [1 * 2 * 3-0 * 1 * 2] + 1 / 3 [2 * 3 * 4-1 * 2 * 3]. 1 / 3 [19 * 20 * 21-18 * 19 * 20] = 1 / 3 [1 * 2 * 3-1 * 2 * 3 + 2 * 3 * 4. - 18 * 19 * 20 + 18 * 19 * 20] = 1 / 3 [19 * 20 * 21] = 2660, the middle part is about 2) 1 × 2 + 2 × 3 +. + a

Reading materials: when he was in school, the great mathematician Gauss once studied such a problem: 1 + 2 + 3 + 4 + 5 +... + 100 =? After

（1+99）+（2+98）+（3+97）+（4+96）.+（49+51）+50+100=100+100+100...+100+50+100
There are 49 + 1 100 and 1 50
==49*100+50+100
=5500

Reading material, mathematician Gauss once studied such a problem in school

(1)343400
(2)
1×2=（1×2×3-0×1×2）/3
2×3=（2×3×4-1×2×3）/3
3×4=（3×4×5-2×3×4）/3
.
n(n+1)=(n(n+1)(n+2)-(n-1)n(n+1))/3
So 1 × 2 + 2 × 3 + + n(n+1)=n(n+1)(n+2)/3

Given that a + b > 0, prove that a ^ 3-A ^ 2b is greater than AB ^ 2-B ^ 3