Math Problem Statement

Find the minimum value of : \frac{{{{\log }a}\left( {bc} \right) + {{\log }b}\left( {ac} \right) + {{\log }c}\left( {ab} \right)}}{{{{\log }{ab}}c + {{\log }{bc}}a + {{\log }{ca}}b}} where a, b, c ∈ (0, 1) ∪ (1, ∞).

Solution

We need to find the minimum value of the expression:

loga(bc)+logb(ac)+logc(ab)logabc+logbca+logcab\frac{{\log_a(bc) + \log_b(ac) + \log_c(ab)}}{{\log_{ab}c + \log_{bc}a + \log_{ca}b}}

Step 1: Simplify the Numerator

We start by simplifying the numerator term by term using the logarithmic identity:

logx(yz)=logx(y)+logx(z)\log_x(yz) = \log_x(y) + \log_x(z)

So:

loga(bc)=loga(b)+loga(c)\log_a(bc) = \log_a(b) + \log_a(c) logb(ac)=logb(a)+logb(c)\log_b(ac) = \log_b(a) + \log_b(c) logc(ab)=logc(a)+logc(b)\log_c(ab) = \log_c(a) + \log_c(b)

Therefore, the numerator becomes:

loga(b)+loga(c)+logb(a)+logb(c)+logc(a)+logc(b)\log_a(b) + \log_a(c) + \log_b(a) + \log_b(c) + \log_c(a) + \log_c(b)

Step 2: Simplify the Denominator

We now simplify the denominator using the logarithmic identity:

logxyz=logzlog(xy)=logzlogx+logy\log_{xy}z = \frac{\log z}{\log (xy)} = \frac{\log z}{\log x + \log y}

So:

logabc=logcloga+logb\log_{ab}c = \frac{\log c}{\log a + \log b} logbca=logalogb+logc\log_{bc}a = \frac{\log a}{\log b + \log c} logcab=logblogc+loga\log_{ca}b = \frac{\log b}{\log c + \log a}

Therefore, the denominator becomes:

logcloga+logb+logalogb+logc+logblogc+loga\frac{\log c}{\log a + \log b} + \frac{\log a}{\log b + \log c} + \frac{\log b}{\log c + \log a}

Step 3: Consider Symmetry and Apply the AM-GM Inequality

Let’s assume a=b=ca = b = c. In this case, the numerator simplifies to:

loga(b)+loga(c)+logb(a)+logb(c)+logc(a)+logc(b)=6loga(a)=6×0=0\log_a(b) + \log_a(c) + \log_b(a) + \log_b(c) + \log_c(a) + \log_c(b) = 6 \log_a(a) = 6 \times 0 = 0

This approach seems unsuitable, so we look at another symmetrical scenario. Consider using the AM-GM inequality in the denominator.

The expression for the denominator can be handled by the inequality:

logcloga+logb+logalogb+logc+logblogc+loga3logalogblogc3loga+logb+logc3=3\frac{\log c}{\log a + \log b} + \frac{\log a}{\log b + \log c} + \frac{\log b}{\log c + \log a} \geq \frac{3\sqrt[3]{\log a \log b \log c}}{\frac{\log a + \log b + \log c}{3}} = 3

Given that this is a cyclic sum, the minimum possible value of this expression considering AM-GM is 1.

If a=b=ca = b = c, the minimum value is achieved and equals to 2.

So, if a=b=ca = b = c, we conclude that both the numerator and the denominator simplify to the same form, leading to a minimum value of:

Minimum value=2\text{Minimum value} = 2

Thus, the minimum value of the expression is 2.

Would you like to explore the specific steps or assumptions made in this solution? Or perhaps a different problem?

Here are five questions related to this topic:

  1. How does the AM-GM inequality apply to logarithmic functions?
  2. What happens if aa, bb, and cc are distinct but close in value?
  3. Can the minimum value of the expression change if aa, bb, and cc belong to different ranges?
  4. What are the implications of different logarithmic bases for such an expression?
  5. How would you handle the expression if logarithms with different bases were not allowed?

Tip: Always check for symmetry when solving expressions with multiple variables—it can often simplify the problem significantly.

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Math Problem Analysis

Mathematical Concepts

Logarithms
AM-GM Inequality

Formulas

Logarithmic identities
AM-GM inequality

Theorems

Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality)

Suitable Grade Level

Advanced High School / College