Alien Inversion?

$\tan^{-1}\sqrt{a}+\tan^{-1}\sqrt{b}+\tan^{-1}\sqrt{c}=\dfrac{\pi}{2}$ $\implies \sqrt{ab}+\sqrt{bc}+\sqrt{ca}=1$

$\sin^{-1}x+\sin^{-1}y+\sin^{-1}z=\dfrac{\pi}{2}$ $\implies x^2+y^2+z^2=1-2xyz$

From Titu's Lemma , we have $\dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}\geq \dfrac{(a+b+c)^2}{x+y+z}$ .

So for $\dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}$ to be minimum, $(a+b+c)$ should be minimum and $(x+y+z)$ has to be maximum.

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Consider $\{\sqrt{a},\sqrt{b},\sqrt{c}\} \text{and} \{\sqrt{b},\sqrt{c},\sqrt{a}\}$

Applying Cauchy-Schwarz Inequality we get,

$(a+b+c)^2\geq(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})^2$

$\implies (a+b+c)\geq 1$

Therefore, minimum value of $(a+b+c)$ is $\boxed{1}$ .

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Consider $\{x,y,z\} \text{and} \{1,1,1\}$

Applying Cauchy-Schwarz Inequality we get,

$3(x^2+y^2+z^2)\geq(x+y+z)^2$

$3(1-2xyz)\geq(x+y+z)^2$

But given that $xyz=\dfrac 18$

So we obtain $(x+y+z)^2\leq3(1-\frac 28)$

$\implies$ $(x+y+z)^2\leq3(1-\frac 14)$

$\implies (x+y+z)\leq\dfrac 32$

Therefore, maximum value of $(x+y+z)$ is $\boxed{\dfrac 32}$

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Finally substituting the two obtained values in the required expression, we get

$\dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}\geq\dfrac{(1)^2}{\frac 32}$

$\implies \dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}\geq\dfrac{2}{3}$

Hence, the minimum value of $\dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}$ is $\boxed{\dfrac{2}{3}}$ .