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Minimum of (a+b)/c + (b+c)/a + (c+a)/b | JEE

JEE Maths question with a full step-by-step solution.

Question

a, b, c

are three positive numbers, then the minimum value of

\frac{ab(a+b)+bc(b+c)+ca(c+a)}{abc}

3

4

6

correct

1

Solution

Step 1: Divide each term by

abc

\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}=\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}.

Step 2: AM–GM on the six ratios, whose product is

1

\frac{\text{sum}}{6}\ge\left(\frac{a}{c}\cdot\frac{b}{c}\cdot\frac{b}{a}\cdot\frac{c}{a}\cdot\frac{c}{b}\cdot\frac{a}{b}\right)^{1/6}=1.

Step 3: So the sum is

\ge 6

, with equality when

a=b=c.

Correct answer: (3)

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