Given that (a + c)(b + d) = ac + bd
Problem Statement:
Let \(a, b, c, d\) be positive real numbers satisfying:
\((a + c)(b + d) = ac + bd\).
We want to find the minimum value of the expression:
\(\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a}\).
Expanding the left-hand side:
\((a + c)(b + d) = ab + ad + cb + cd\)
= \(ab + ad + bc + cd\)
We are given
\((a + c)(b + d) = ac + bd\),
so we get:
\(ab + ad + bc + cd = ac + bd\)
Bringing all terms to one side:
\(ab + ad + bc + cd – ac – bd = 0\)
Grouping terms:
\((ab + bc) + (ad + cd) – (ac + bd) = 0\)
Let \(a = x\), \(c = x\), \(b = y\), \(d = y\).
Then \((a + c)(b + d) = 4xy\),
and \(ac + bd = x^2 + y^2\),
So: \(4xy = x^2 + y^2\)
\(\Rightarrow x^2 – 4xy + y^2 = 0\)
\(\Rightarrow (x – y)^2 = 2xy\)
Now evaluate:
\(\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a}\)
= \(2\left(\frac{x}{y} + \frac{y}{x}\right)\)
Let \(t = \frac{x}{y}\)
\(\Rightarrow \frac{y}{x} = \frac{1}{t}\),
So: \(2(t + \frac{1}{t})\).
Using earlier condition:
\(4t = t^2 + 1\)
\(\Rightarrow t^2 – 4t + 1 = 0\)
\(\Rightarrow t = 2 \pm \sqrt{3}\)
Then \(t + \frac{1}{t} = 4\)
\(\Rightarrow 2(t + \frac{1}{t}) = 8\).
Final Answer:
Minimum Value:
\[{8}\]
Khairun
