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Algebra Difficulty 4.7 AIME Find the answer

2. If a,b,c>0,1a+2b+3c=1a, b, c > 0, \frac{1}{a}+\frac{2}{b}+\frac{3}{c}=1, then the minimum value of a+2b+3ca+2b+3c is . \qquad

A number or a short expression. Spacing and $ signs are ignored.

Solution

2. 36 .

By Cauchy-Schwarz inequality, we have
a+2b+3c=(a+2b+3c)(1a+2b+3c)⩾(1+2+3)2=36. \begin{array}{l} a+2 b+3 c=(a+2 b+3 c)\left(\frac{1}{a}+\frac{2}{b}+\frac{3}{c}\right) \\ \geqslant(1+2+3)^{2}=36 . \end{array}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.