The minimum value of {left( {frac{3}{a} - 1} | vedclass

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Question

(C) Let $z = {\left( {\frac{3}{a} - 1} \right)^2} + {\left( {\frac{a}{b} - 1} \right)^2} + {\left( {\frac{b}{c} - 1} \right)^2} + {\left( {3c - 1} \ri

Vedclass · Accepted Answer

$27$

Vedclass · Answer

(C) Let $z = {\left( {\frac{3}{a} - 1} \right)^2} + {\left( {\frac{a}{b} - 1} \right)^2} + {\left( {\frac{b}{c} - 1} \right)^2} + {\left( {3c - 1} \right)^2}$.Expanding the terms,we get $z = \left( \frac{9}{a^2} + \frac{a^2}{b^2} + \frac{b^2}{c^2} + 9c^2 \right) + 4 - 2 \left( \frac{3}{a} + \frac{a}{b} + \frac{b}{c} + 3c \right)$.Using the Arithmetic Mean-Geometric Mean $(AM \ge GM)$ inequality:$\frac{1}{4} \left( \frac{9}{a^2} + \frac{a^2}{b^2} + \frac{b^2}{c^2} + 9c^2 \right) \ge \sqrt[4]{\frac{9}{a^2} \cdot \frac{a^2}{b^2} \cdot \frac{b^2}{c^2} \cdot 9c^2} = \sqrt[4]{81} = 3$.So,$\frac{9}{a^2} + \frac{a^2}{b^2} + \frac{b^2}{c^2} + 9c^2 \ge 12$.Similarly,$\frac{1}{4} \left( \frac{3}{a} + \frac{a}{b} + \frac{b}{c} + 3c \right) \ge \sqrt[4]{\frac{3}{a} \cdot \frac{a}{b} \cdot \frac{b}{c} \cdot 3c} = \sqrt[4]{9} = \sqrt{3}$.So,$2 \left( \frac{3}{a} + \frac{a}{b} + \frac{b}{c} + 3c \right) \ge 8\sqrt{3}$.Thus,$z \ge 12 + 4 - 8\sqrt{3} = 16 - 8\sqrt{3}$.Comparing with $p - q\sqrt{r}$,we have $p = 16$,$q = 8$,$r = 3$.Since $q$ and $r$ are coprime,$p + q + r = 16 + 8 + 3 = 27$.

The minimum value of ${\left( {\frac{3}{a} - 1} \right)^2} + {\left( {\frac{a}{b} - 1} \right)^2} + {\left( {\frac{b}{c} - 1} \right)^2} + {\left( {3c - 1} \right)^2}$ where $0 < a, b, c \leqslant 9$,is $p - q\sqrt{r}$; $p, q, r \in I$ and $q, r$ are coprime,then $(p + q + r)$ is equal to

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