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Least Value of a1+2a2+3a3+4a4+5a5, Product 6 | JEE

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

Question
If ai>0a_i>0 for 1≤i≤51\le i\le 5 and a1a2a3a4a5=6a_1 a_2 a_3 a_4 a_5=6, then the least value of a1+2a2+3a3+4a4+5a5a_1+2a_2+3a_3+4a_4+5a_5 is
A6(5!)1/56(5!)^{1/5}
B15(6!)1/515(6!)^{1/5}
C(6!)1/5(6!)^{1/5}
D5(6!)1/55(6!)^{1/5}correct
Solution
Step 1: AM–GM on a1,2a2,3a3,4a4,5a5a_1, 2a_2, 3a_3, 4a_4, 5a_5:
a1+2a2+3a3+4a4+5a55≥(a1⋅2a2⋅3a3⋅4a4⋅5a5)1/5.\frac{a_1+2a_2+3a_3+4a_4+5a_5}{5}\ge\big(a_1\cdot 2a_2\cdot 3a_3\cdot 4a_4\cdot 5a_5\big)^{1/5}.
 Step 2: The product inside is 5! a1a2a3a4a5=120⋅6=720=6!5!\,a_1a_2a_3a_4a_5=120\cdot 6=720=6!:
RHS=(6!)1/5.\text{RHS}=(6!)^{1/5}.
 Step 3: So a1+2a2+3a3+4a4+5a5≥5(6!)1/5.a_1+2a_2+3a_3+4a_4+5a_5\ge 5(6!)^{1/5}. Correct answer: (4)
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