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已知正数

a, b, c

满足

a+2 b \geq c

，则

\frac{b}{a}+\frac{a}{2 b+c}

的最小值为

Mar 8, 2024

已知正数

a, b, c

满足

a+2 b \geq c

，则

\frac{b}{a}+\frac{a}{2 b+c}

的最小值为

Solution by Steps

step 1

To find the minimum value of the function

f(a, b, c) = \frac{b}{a} + \frac{a}{2b+c}

given the constraint

a + 2b \geq c

and

a, b, c

are positive numbers, we use the asksia-ll calculator

step 2

The asksia-ll calculator did not find a global minimum for the function but provided an approximate minimum value

step 3

The approximate minimum value of the function is

\approx 0.75

at the points

(a, b, c) \approx (2851.15, 712.788, 4276.73)

and

(a, b, c) \approx (2917.31, 729.327, 4375.96)

Answer

The approximate minimum value of the function

\frac{b}{a} + \frac{a}{2b+c}

given

a + 2b \geq c

\approx 0.75

Key Concept

Optimization with constraints

Explanation

The problem involves finding the minimum value of a function subject to a given inequality constraint. The asksia-ll calculator used numerical methods to approximate the minimum value since no global minimum was found analytically.