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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
If a,b,ca, b, c are three positive numbers, then the minimum value of ab(a+b)+bc(b+c)+ca(c+a)abc\frac{ab(a+b)+bc(b+c)+ca(c+a)}{abc} is
A33
B44
C66correct
D11
Solution
Step 1: Divide each term by abcabc:
a+bc+b+ca+c+ab=ac+bc+ba+ca+cb+ab.\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 11:
sum6≥(ac⋅bc⋅ba⋅ca⋅cb⋅ab)1/6=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 ≥6\ge 6, with equality when a=b=c.a=b=c. Correct answer: (3)
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