Just make it least

Since $\color{#3D99F6}{a^2 = bc}$ $\implies a+b+c = a\color{#3D99F6}{bc} = a\cdot \color{#3D99F6}{a^2} = a^3$ . Then $ab+\color{#3D99F6}{bc}+ca$ $= ab+\color{#3D99F6}{aa}+ca$ $=a(b+a+c) = a\cdot a^3 = a^4$ . Therefore, $\color{#3D99F6}{a^4} + \color{#D61F06}{a^2} = \color{#3D99F6}{ab+bc+ca} + \color{#D61F06}{bc}$ . Since $a, b, c > 0$ , we can use AM-GM inequality as follows:

$\begin{aligned} ab+bc+ca+bc & \ge 4\sqrt[4]{a^2b^3c^3} \\ a^4 + a^2 & \ge 4\sqrt[4]{a^8} = 4a^2 \\ \implies a^4 & \ge 3a^2 \\ a^2 & \ge 3 \\ a^4 & \ge 9 \end{aligned}$

$\begin{aligned} \implies a^4 + a^2 - 7 & \ge 9 + 3 -7 = \boxed{5} \end{aligned}$

By AM-GM, we have

$a + b + c \ge 3\sqrt [ 3 ]{ abc }$ .

Squaring and cancellation alongwith the fact $a + b + c = abc$ , gives

${ \left( abc \right) }^{ 2 }\ge 27$ .

Putting $a^2 = bc$ gives

$bc\ge 3$ or $a^2 \ge 3$ .

So, $a^4 + a^2 -7 \ge { \left( 3 \right) }^{ 2 } + { \left( 3 \right) }^{ 1 } - 7 = \boxed{5}$ .

We have $b,c$ must be the roots of the quadratic $X^2 - (a^3 - a)X + a^2 \; = \; 0$ For this quadratic to have two (possibly repeated) positive roots we need $a^3 - a > 0$ and also $(a^3 - a)^2 - 4a^2 \ge 0$ , so that we must have $a > 1$ and $0 \le a^6 - 2a^4 - 3a^2 = a^2(a^2+1)(a^2-3)$ and hence we must have $a^2 \ge 3$ . This makes the answer $9+3-7=\boxed{5}$ .

Given that $a^2=bc,a+b+c=abc$ Second equation implies, $b+c=a(bc-1)$ Applying AM-GM and substituting value of $bc$ from first equation at the top,we have, $\frac{a(a^2-1)}{2}>=a\Rightarrow a^2 >= 3$ Putting the minimum value we got of $a^2$ we get answer as $\boxed{5}$ .

Since we have $a^2 = bc$ , we can write $b$ and $c$ parametrically as $b = ka$ and $c = \frac{a}{k}$ for some variable real positive constant $k$ (as $a,b,c > 0$ . Substituting this into $a+b+c = abc$ , we get $a + ka + \frac{a}{k} = a(ka)(\frac{a}{k}) = a^3$ . Since $a > 0$ , we must have $1 + k + \frac{1}{k} = a^2$ . Therefore for any $k$ , we can generate a value for $a$ and thus values for $b$ and $c$ .

To minimise $a^4 + a^2 - 7$ , this is a strictly increasing function in $a$ for $a > 0$ so to find the minimum of this, we need the minimum value for $a$ (or more over, $a^2$ ). Since we can make $a^2$ as small as possible by making $1 + k + \frac{1}{k}$ as small possible too, using the AM-GM inequality, we have $1 + k + \frac{1}{k} \geq 3\sqrt{1\times k \times \frac{1}{k}} = 3$ where equality holds when $k = 1$ . This means the least value of $a^2$ is $3$ and futher $a^4$ is $3^2 = 9$ . Therefore the smallest value of $a^4 + a^2 - 7$ is $9 + 3 - 7 = \boxed{5}$ .