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对任意的正实数

a, b ， c

，满足

b+c=1

, 则

\frac{3 a b^{2}+a}{b c}+\frac{12}{a+1}

的最小值为

Mar 8, 2024

对任意的正实数

a, b ， c

，满足

b+c=1

, 则

\frac{3 a b^{2}+a}{b c}+\frac{12}{a+1}

的最小值为

Solution by Steps

step 1

Given the function

f(a, b, c) = \frac{3ab^2 + a}{bc} + \frac{12}{a+1}

with the constraint

b + c = 1

, we are asked to find its minimum value for positive real numbers

a, b, c

step 2

Substitute

c = 1 - b

into the function due to the constraint

b + c = 1

step 3

The function simplifies to

f(a, b) = \frac{3ab^2 + a}{b(1 - b)} + \frac{12}{a+1}

step 4

The asksia-ll calculator did not find a global minimum for the function

Answer

The function does not have a global minimum.

Key Concept

Optimization with Constraints

Explanation

When optimizing a function with constraints, substitution is often used to reduce the number of variables. However, not all functions have a global minimum, especially in an unbounded domain for positive real numbers.