What is the minimum value of the algebraic formula a 2 1 b a b if a b 0?

Application of the mean value theorem.

Because a>b>0 , b(a-b)<= 2=a 2 4 , so a 2+1 [b(a-b)]>=a 2+4 a 2>=2* (a 2*4 a 2)=4 , when b=a-b and a 2=4 a 2 i.e. a = 2, b= 2 2, the minimum value is 4.

Because a 2+1 b(a-b)>=2*open root number [a 2 b*(a-b)] The condition for taking the minimum value is: a 2 = 1 b * (a-b) ......1）

Therefore, this problem is equivalent to finding the minimum value of a 2 b*(a-b).

And because a=b+(a-b).

Therefore, a 2 b*(a-b)=[b+(a-b)] 2 b*(a-b)>=4*b*(a-b) b*(a-b)=4 The condition for removing the minimum value is b=(a-b)......2）

Combining the two equations of 1 and 2 gives a = open root number 2 b = open number 2 2, that is, there is such a number, but the inequality exists to remove the minimum value.

then its minimum value is.

a 2+1 b(a-b)>=2*open root number[a 2 b*(a-b)] =2*open root number[b+(a-b)] 2 b*(a-b)>=2*open root number4*b*(a-b) b*(a-b)=2*open root number4=2*2=4

It's better to ask the math teacher ...... directlyYou're not going to be in the exam...... right

Since 1 a+4 b = 1, then a+b = (a + b) * (1 a + 4 b) = 5 + b a + 4a b> = 5 + 2 * under the root number ((b a)*(4a b)) = 9, if and only if b a = 4a b, by 1 a + 4 b, the simultaneous solution a = 3, b = 6

Mean inequality is used quadratically.

by a>b>0

a-b>0

So b*(a-b)<=(b+a-b) 2 4=a 2 4 (if and only if b=a-b, i.e., a=2b, etc.).

So 1 b*(a-b)>=4 a2

So, a 2+1 b*(a-b).

a^2+4/a^2

If and only if a 2 = 4 a 2 i.e. a = root number 2, etc.) the minimum value of the total a 2 + 1 b * (a-b) 4, obtained when a = root number 2, b = root number 2 2.

2(1/a+4/b)

a+b)(1 Senxuxiana+4 b).

5+(4a/b+b/a)

4a b+b a 2 (4a this beat b*b a) = 42 (1 a + 4 b) 4 + 5 = 9

The minimum value before the reputation is 9 2

1\a+1\b=（a+b）/ab=2/aba＞0，b＞0

So A+B 2 root number (ab).

2 root number (ab) quietly.

Root No. Piyun Jane AB 1

0＜ab≤1

So. 1 a+1 b=(a+b) ab=2 ab 2 1=2So. The minimum burn value of 1 A+1 B is 2, and there is no maximum value.

Application of the mean value theorem.

Because a>b>0

So b(a-b)<= 2=a 2 4, so a 2+1 [b(a-b)]>=a 2+4 a 2>=2* (a 2*4 a 2)=4, when.

b = a-b and a 2 = 4 a 2

i.e. a = 2, b = 2 2

, the minimum value is 4.

Solution: Because x>a>0, then x2 (x-1)=[(x-a) 2+2a(x-a)+a 2] (x-a).

x-a)+[a 2 (x-a)]+2a>=2a+2*root.

2a+2a=4a, so this.

The minimum value of an algebraic formula is 4a

I don't agree with the answer on the second floor, when taking the equals sign, we should make sure that 2b a = 16a b and b 2 a 2 = 8a 2 b 2 are valid at the same time, that is, b 2 = 8a 2 and b 4 = 8a 4 are valid at the same time, and the latter b 2 = 2 2a 2, so it is not true.

The correct solution is as follows:

According to the title, let a=x, b=1-x

0 let 1 a 2+8 b 2=y, then y=1 x 2+8 (1-x) 20y'=-2/x^3+16/(1-x)^3

16x 3-2(1-x) 3] x 3(1-x) 3 reams'=0, then 8x 3=(1-x) 3, 2x=1-x, and x=1 3 shows that the function y has only one extreme point, and x cannot be equal to the value of both ends in its value range.

So when x=1 3, the function y takes the minimum value =27

That is, the minimum value of 1 a 2 + 8 b 2 is 27

a+b=12/a

1/b2(a+b)/a

a+b)/b

2b/aa/b

a>0b>0a/b>0

b/a>0

From the mean inequality, when 2b a

a b, i.e., a = 2b, 2b a

A b has a minimum value of 2 2

At this point, 2 a1 b has a minimum value of 3 + 2 2.