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

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Question

(C) Let $z = \left(\frac{3}{a}-1\right)^2+\left(\frac{a}{b}-1\right)^2+\left(\frac{b}{c}-1\right)^2+(3c-1)^2$.
For minimization, each term is non-neg

Vedclass · Accepted Answer

$27$

Vedclass · Answer

(C) Let $z = \left(\frac{3}{a}-1\right)^2+\left(\frac{a}{b}-1\right)^2+\left(\frac{b}{c}-1\right)^2+(3c-1)^2$.
For minimization, each term is non-negative, so the minimum occurs when all expressions are equal:
$\frac{3}{a} = \frac{a}{b} = \frac{b}{c} = 3c = k$
From $3c = k$, we get:
$c = \frac{k}{3}$
From $\frac{b}{c} = k$, we get:
$b = kc = \frac{k^2}{3}$
From $\frac{a}{b} = k$, we get:
$a = kb = \frac{k^3}{3}$
From $\frac{3}{a} = k$, we get:
$3 = ak = \frac{k^4}{3}$
$k^4 = 9 \Rightarrow k = \sqrt{3}$
Now substituting $k = \sqrt{3}$:
Each term becomes:
$\sqrt{3} - 1$
So,
$z = 4(\sqrt{3} - 1)^2$
$= 4(3 + 1 - 2\sqrt{3})$
$= 16 - 8\sqrt{3}$
Comparing with $p - q\sqrt{r}$, we get:
$p = 16,\ q = 8,\ r = 3$
Thus,
$p + q + r = 16 + 8 + 3 = 27$

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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