Let a 0 and b 0, and if the root number 3 is the middle term of 3 a and 3 b, then the minimum value

Since the root number 3 is the proportional middle term of 3 a and 3 b, (3) = 3 a*3 b, so 3 (a+b) = 3

That is, a+b=1, which can be solved according to a 0, b 0, using the basic inequality.

1/a+2/b=(a+b)/a+2(a+b)/b=1+b/a+2a/b+2≥2√(b/a*2a/b)+3=2√2+3

The minimum value of 1 a+2 b is 2 2+3

The root number 3 is the proportional middle term of 3 a and 3 b.

3=3 (a+b) power, i.e., a+b=1

1 a+2 b = (a + b) (1 a + 2 b) = 1 + 2 + 2a b + b a 3 + 2 root number 2

Since a 0, b 0 , and the root number 3 are the proportional middle terms of 3 a and 3 b, a multiplied by b equals the square of 3 which is 9

So a=9 b, so 1 a+2 b=b 9+2 b, and then the mean inequality is used, so b 9+2 b>=2 multiplied by the root number 2 and divided by 3 is the minimum.

3^b=3*3^a b=a+1

1 a+2 b=1 a+2 (a+1) when 1 a=2 (a+1) and minimum, a=1, and minimum is 2

Since the root number 3 is the proportional middle term of 3 a and 3 b, then 3 a*3 b = 3, then a+b = 1, and a 0, b 0, then 1 a+2 b=(a+b) a+2*(a+b) b=

1+b a+2*(1+a b)=3+a b+2*b a>=3+2 times root number 2

3=3^a * 3^b

a+b=11/a+2/b=(a+b)/a +(2a+2b)/b1+b/a+ 2a/b +2

The equal sign holds if and only if b a= 2a b.

The minimum value is 3+2 2

i.e., 3 is the proportional middle term of 3 a and 3 b.

3^a x 3^b=3

a+b=11/a+1/b

a+b)/a+(a+b)/b

2+b a+a b 2+2 (b axa b)=2+2=4 if and only if b a=a b, i.e., a=b=1 2, the equal sign is established.

Therefore, the minimum value of 1 a+1 b is 4

It can be obtained from the title.

3^(a+b)=3

So a+b=1

1 a+1 b = (a+b) ab

A>0, B>0, so A+B>=2 Ab so 0

√3b)^2=(1-a)(1+a)

3b^2=1-a^2

a^2+3b^2=1

Let a=sint,b=cost 3

then a+3b=sin t + 3cost

2(1/2*sint +√3cost）

2(sint*cosπ/3 +cost*sinπ/3）2sin(t+π/3)

1<=sin(t+π/3)<=1

So the maximum value of A+3B is: 2

I'll offer you an idea that you can do the math and see. If there is a condition, we can get: a 2+3b 2=1, so we can set a = sinx, (root number 3) b = cosx, and then substitute it into a + 3b, note -1 "a "1, - (root number 3) 3 "b " (root number 3).

Find the maximum value of the trigonometric function.

√3b)^2=(1-a)(1+a)

3b^2=1-a^2

a^2+3b^2=1

x= 3b y=a x 2+y 2=1 a+3b=z=y+ 3xx 2+y 2=1 a+3b=z=y+ 3x is the equation for the sum of straight lines.

So d=2 absolute value = <1 absolute value z=《2 so the maximum value is 2

The root number 3 is the proportional middle term 3 (a+b) = (root number 3) =3 a + b = 1 a +1 b = (a+b) ab =1 ab ab (a+b) 4 =1 4 so 1 a +1 b =(a+b) ab =1 ab 4 so the minimum value is 4

(√3b)^2=(1-a)(1+a)

3b^2=1-a^2

a^2+3b^2=1

Let a=sint,b=cost 3

then a+3b=sin t + 3cost

2(1/2*sint +√3cost）

2(sint*cosπ/3 +cost*sinπ/3）=2sin(t+π/3)

1<=sin(t+π/3)<=1

So the maximum value of A+3B is: 2

9=3 a*3 b gives a+b=2 1 a+1 b=(a+b) ab=2 ab because a+b》2 root number ab

a=b when ab is established max 2 ab min = 2 1 = 2

Solution: Since 3 is the proportional middle term of 3 a and 3 b, so (3) 2=3 a*3 b=3 (a+b), 3=3 (a+b), so a+b=1and a>0, b>0, so a+b>=2 ab, i.e., ab<=1 4,1 ab>=4

So 1 a+1 b=(a+b) ab=1 ab>=4So the minimum value of 1 a+1 b is 4.

Because (root number 3) is the middle term of 3 a in proportion to 3.

So 3 a*3 =3 we know a+b=1

1 a+1 b=(a+b) ab=1 ab requires the minimum value of 1 a+1 b i.e., find the maximum value of a*b and the maximum value of a*b is.

So the minimum value of 1 a+1 b is 4

Because the root number 3 is the proportional middle term between the power of 3 a and the power b of 3, 3 a*3 b = 3, that is, a + b = 1, 1 a + 1 b = (a + b) a + (a + b) a = 2 + b a + a b Since a > b, b > 0,2 + b a + a b> = 2 + 2 sqrt (a b * b a) = 4, the equal sign is a = b = 1 2

So the suspicion condition is a>=b,b>0

That minimum value is 4, otherwise there is no minimum value.