Knowing a 0, b 0 5 a 3 a, c 0 5 b 3 b, try to compare the size of a, b, c 50

Knowing a>0, b=, c=, try to compare the size of a, b, c.

Solution: a>0, b=, c=

b>0,c>0

1) Set a function.

f(x)=0) if f(x)=0, x= 3, if f(x)>0,x< 3 if f(x)<0,x> 3

2) From above, compare a and b first

When a> 3, b-a<0, ba

When a=3, b=a

Compare b, c

a>0b=>=√3

c<=b

To sum up the above:

a=3, a=b=c

a=3, a=b=c

A< 3, B>A, B>C

There is a problem pm

a+3 a>=2 root number 3 and only if a 2=3 is true.

So b> = root number 3

b+3 b>=2 root number 3 and only true if b 2=3.

So c> = root number 3

a=b=cOther times a

It is not possible to compare, and the conditions are still poor, if you take a=1, then b=2, c= b>c>a

Again, if a=4, then we get b= c=

then there are a>b>c

Method 1: a>0, b>0, and.

2a+3b＝ab→a＝3b/(b-2).

a+2b＝3b/(b-2)+2b

2(b-2)+6/(b-2)+7

2 [2(b-2)·6 (b-2)]+7, so the minimum value is 4 3+7

Method 2: 2a+3b ab

1＝3/a+2/b

√3)^2/a+2^2/(2b)

√3+2)^2/(a+2b),∴a+2b≥(√3+2)^2＝7+4√3.

Therefore, the minimum value is 7+4 3.

a^2=2 b^3=3

a^2)^3=a^6=2^3=8

b^3)^2=b^6=3^2=9

So aca 2=2 c 5=5

a^2)^5=a^10=2^5=32

c^5)^2=c^10=5^2=25

So a>c

The training is noisy. B> Touch A in Zen > congratulate C

Since the positions of a and b are equal, it can be rotated to set a>b without affecting the judgment of the result.

Then: a a*b b a b*b a

a^b*b^b*(a^(a-b))-a^b*b^b*(b^(a-b))

a b*b b*(a (a-b)-b (a-b)) can be seen because a>b, so a (a-b)-b (a-b)>0 so a a*b b>a b*b a

Classmate, let me teach you, you see, this kind of question is a and b, both have positive and negative numbers, at first glance, it is very clueless. Today I will teach you an important idea classification solution in mathematics. You see we catch a, you see situation 1 when a = 5, b4 or minus 4 do not meet a + b < 0, exclude.

Case 2, when a=-5, b=-4 or 4 all match a+b<0, so it can only be a=-5, b=4 or a=-5, b=-4, so the answer is -9 or -1

Solution: <>

1) When a>b>0, <>

Thus the function <>

Increasing. <>

2) When b>a>00, 0 <>

1, a-b 0, thus the function <>

Degression. <>

3) When a=b, abbbab

BA To sum up, we can know <>

y=1/a+4/b

1/2*2*(1/a+4/b)

1/2*(a+b)*(1/a+4/b)

1/2*[（a）^2+(√b)^2]*≥1/2*^2

according to Cauchy's inequality).

So the minimum value of y is 9 2

The following answers are all wrong, and the Cauchy inequality, which can be solved with the basic inequality:

are all positive numbers, so a+b=2>=2 times the root number ab

So the maximum value of the root number ab is 1

So y=1 a+4 b>=2 times the root number 4 ab

So y> = 4/4 of the root number abWhen the denominator is the largest, y takes the minimum value. The maximum value of the denominator is 1

So the minimum value of y is 4