Squared and reversed

Similar solution to @Harsh Shrivastava , perhaps better presented.

By Cauchy-Schwarz inequality :

$\begin{aligned} \left(\color{#D61F06}{\frac 1{\sqrt 3}}\color{#3D99F6}{\sqrt 3 a} + \color{#D61F06}{\sqrt 2}\color{#3D99F6}{\sqrt 2 b} + \color{#D61F06}{3}\color{#3D99F6}{c}\right)^2 & \le \left( \left(\color{#D61F06}{\frac 1{\sqrt 3}}\right)^2 + (\color{#D61F06}{\sqrt 2})^2 + (\color{#D61F06}{3})^2 \right) \left( (\color{#3D99F6}{\sqrt 3 a})^2 + (\color{#3D99F6}{\sqrt 2 b})^2 + (\color{#3D99F6}{c})^2 \right) \\ (a+2b+3c)^2 & \le \left(\frac 13 + 2 + 9 \right)(3a^2+2b^2+c^2) \\ \left( \frac 12 \right)^2 & \le \frac {34}3 (3a^2+2b^2+c^2) \\ \implies 3a^2+2b^2+c^2 & \ge \frac 3{4(34)} \approx \boxed{0.02206} \end{aligned}$

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By Lagrange multipliers :

The problem can also be solved using Lagrange multipliers.

Let $F(a,b,c,\lambda) = 3a^2+2b^2+c^2 - \lambda \left(a+2b+3c - \dfrac 12\right)$ , where $\lambda$ is the Lagrange multiplier.

Then, we have: $\dfrac {\partial F}{\partial a} = 6a - \lambda$ , $\dfrac {\partial F}{\partial b} = 4b - 2\lambda$ , $\dfrac {\partial F}{\partial c} = 2c - 3 \lambda$ and $\dfrac {\partial F}{\partial \lambda} = a+2b+3c - \dfrac 12$ .

Equating $\dfrac {\partial F}{\partial a} = \dfrac {\partial F}{\partial b} = \dfrac {\partial F}{\partial c} = \dfrac {\partial F}{\partial \lambda} = 0$ . we have: $6a=2a=\dfrac 23 c = \lambda$ $\implies b=3a$ and $c=9a$ . Substituting in $a+2b+3c = \dfrac 12$ , we have $a = \dfrac 1{68}$ , and substituting that in $3a^2+2b^2+c^2$ we get the minimum. $\implies 3a^2+2b^2+c^2 = (3 + 18+81)a^2 = \dfrac {102}{68^2} \approx \boxed{0.02206}$

By Cauchy Schwarz Inequality ,

$((\sqrt{3}a)^2+(\sqrt{2}b)^2 + (c)^{2})((\frac{1}{\sqrt{3}})^2 + (\sqrt{2})^2 +(3)^2) \geq (a+2b+3c)^2$

$\implies$ Minimum value of our expression is $0.02205882352$ .