Knowing that a, b, and c are all positive real numbers, then a b c 1 a b 1 c the minimum value is

Let m = a+b, and it is clear that m is a positive real number, then the equation evolves into (m+c)(1 m+1 c)=(m+c) 2 (mc).

I believe you will understand at a glance at this point, the minimum value is obtained when m=a+b=c, and the minimum value is equal to 4.

If you don't understand, please feel free to ask again.

a+b+c)【1/(a+b)+1/c】=（a+b+c)/(a+b)+（a+b+c)/c=1+c/(a+b)+1+（a+b)/c=2+c/(a+b)+（a+b)/c=2+【c²+（a+b)²/c(a+b)】

And the sum of the squares of any two numbers is greater than twice the product of these two numbers (a+b can be seen as a whole) - according to the perfect square.

a+b) = a +b 2ab, and the square of any number is greater than or equal to zero, so (a+b) 0, a +b 2ab 0, a +b 2ab, then a +b 2ab 1, a +b ab 2

Therefore, the minimum value of c + (a+b) c(a+b) 2 and c +(a+b) c(a+b) is 2

a+b+c) [1 (a+b)+1 c] has a minimum value of 2+2=4

Knowing that a, b, and c are all positive, and a+b+c=1, what is the minimum value of 1 a+1 b+1 c? （a+b+c)*(1/a+1/b+1/c) =3+(b/a+a/b+a/c+c/a+b/c

It is obtained by the positive numbers a, b, and c in the mu.

a+b)(b+c)

ab+b^2+bc+ac

b(a+b+c)+ac

b*[1 Mt. Xunzi (ABC)]+AC

ac+1/(ac)

a+b)(b+c) let the smallest value = 2

a+b+c=0;abc=16;If c>0 obtains a, b is less than imitation 0c=-a-b>=2*(-a-b), and a=b when the equal sign is true;

abc>=ab*[2*(-a-b)^;AB=4, and A=B, so A=B=-2

The round celery is big with orange: c = 4

Because a, b, and c are all positive real numbers, and there are.

a＋b＋c＝abc。

Then: abc a b c 3(abc) 1 3, ie.

abc) Tan Yan Na 3 27 abc, abc) 2 27, jujube big abc 3 3.

So what you get is:

abc)min with 3 3.

From a, b, are positive real numbers, and a+b=1 gives ab<=1 4 original formula = ab + 1 (ab) + (a b + b a) = ab + 1 (ab) + (a 2 + b 2) (ab) = ab + 1 (ab) + (a 2 + b 2 + 2 ab) (ab)-2

ab+1/(ab)+(a+b)^2/(ab)-2=ab+1/(ab)+1/(ab)-2=ab+2/(ab)-2

At f(x)=x+2 x, monotonically decreasing at (0, root number 2), so when ab=1 4.

The original takes the minimum value = 25 4

Solution: a+b+c=0; abc=16;If c>0 gets a, b is less than 0c=-a-b>=2*(-a-b), and the equal sign is a=b;

abc>=ab*[2*(-a-b)^;AB=4, and A=B, so A=B=-2

So: c=4

a+b=-c, ab=16 c, (a+b) is perfectly squared -2ab=a squared + b squared is greater than or equal to 0, so the square of c is -32 c is greater than or equal to zero, and multiply it by c to get (c is a positive number, unchanged sign): the cubic of c -32 is greater than or equal to 0, and the solution c is greater than or equal to the cubic root number 32, so the minimum value of c is the cubic root number 32, not 4

a+b=-c ab=16 c a, b is the root of the equation x 2+cx-16 c=0, so c 2-64 c>=0 c 3-64>=0 c 3>=64 c>=4 so c minimum is 4

Solution: 1 a+9 b=1

a+b=（a+b)*(1/a+9/b)=1+9+b/a+9a/b10+b/a+9a/b>=10+2*sqrt(b/a*9a/b)=10+2*3=16

If and only in the group, when b a=9a b, i.e., b=3a, bring in 1 a+9 b=1, then a=4 can be obtained

b=12, so the minimum value of the medium module is 16, sell or slow and only take the minimum value when a=4, b=12.