Question

If \(\log _{2}(a+b)+\log _{2}(c+d) \geq 4\). Then the minimum value of the expression \(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}\) is (1) 2 (2) 4 (3) 8 (4) none of these

Short answer

The minimum value of the expression \(a + b + c + d\) is 8.

Step-by-step solution

Understand the Logarithmic Inequality

Given: \(\log_{2}(a+b) + \log_{2}(c+d) \geq 4\). Since both logarithms are base 2, we combine them using the property of logarithms: \(\log_{2}(a+b) + \log_{2}(c+d) = \log_{2}((a+b)(c+d))\). Thus, the inequality becomes \(\log_{2}((a+b)(c+d)) \geq 4\).

Simplify the Inequality

The inequality \(\log_{2}((a+b)(c+d)) \geq 4\) implies: \( (a+b)(c+d) \geq 2^{4}\). Therefore, \( (a+b)(c+d) \geq 16\).

Analyze Possible Values

We are searching for the minimum value of the expression \(a + b + c + d\). Notice that the minimum occurs when \( (a+b) \) and \( (c+d) \) multiply to 16 with minimum individual sums.

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

By AM-GM inequality, \(a + b \geq 2\sqrt{ab} \) and \(c + d \geq 2\sqrt{cd} \). To achieve the equality, choose \(a = b\) and \(c = d\), thus we get \(a + b = 2a\) and \(c + d = 2c\).

Solve for Minimum Values

Set \(2a \times 2c = 16 \) which simplifies to \( 4ac = 16 \) or \( ac = 4 \). Then we need \(a + a + c + c = 4(a+c) \), where \(a = 1 \) and \(c=1 \) gives \(a+b+c+d = 4(1+1) = 8 \).

Key concepts

Logarithms

Logarithms are a way to express numbers as exponents of a base. They help us solve for unknowns in exponential relationships. For instance, \(\text{log}_{b}(x) = y\) means \( b^y = x \).
This powerful tool is useful for simplifying complex equations involving multiplication and exponentiation.
In the given problem, we use the property of logarithms: \( \text{log}_{b}(x) + \text{log}_{b}(y) = \text{log}_{b}(xy) \). This allows us to combine two logarithms into one, making it easier to handle inequalities.
For example, \( \text{log}_{2}(a+b) + \text{log}_{2}(c+d) \) simplifies to \( \text{log}_{2}((a+b)(c+d)) \).

Arithmetic Mean-Geometric Mean Inequality

The Arithmetic Mean-Geometric Mean (AM-GM) Inequality is a fundamental principle in mathematics. It states that for any non-negative real numbers, the arithmetic mean is always greater than or equal to the geometric mean.
Mathematically, it is expressed as: \[ \frac{a + b}{2} \geq \sqrt{ab} \]
Equality holds if and only if \(a = b\).
In our solution, we apply AM-GM inequality to find the minimum sums of \( (a+b) \) and \( (c+d) \). This is done by setting \(a = b\) and \(c = d\), which simplifies our expressions and helps us find values that satisfy the given condition, \((a+b)(c+d) \geq 16 \).

Minimum Value Analysis

Minimum value analysis helps us find the smallest possible value of an expression that satisfies given conditions. In this problem, we need to find the minimum value of \(a + b + c + d\).
First, we solve the inequality \((a+b)(c+d) \geq 16\). Using the equality case of the AM-GM inequality, we set \((a+b) = (c+d) = 4\).
This minimizes our individual sums, ensuring conditions are met with minimal values. Solving for \(a = b \) and \(c = d\), we find that each pair must equal 4. Thus, \(a + b + c + d = 4 + 4 = 8\), giving us the minimum value of the expression as 8.