For any prime $p$, let $p_a$ be its exponent in the prime factorisation of $a$, and similarily define $p_b, p_c$. Thus, for instance, $a = 2^{2_a}\cdot 3^{3_a}\cdot 5^{5_a}\cdots$, where most of these factors are $1$.
Now consider the exponent of $p$ in the prime factorisations of the left and right hand sides. On the left-hand side it's $\max(p_a, \min(p_b, p_c))$ and on the right-hand side it's $\min(\max(p_a, p_b), \max(p_a, p_c))$. We see that if $p_a<p_b, p_c$, then we get $\min(p_b, p_c)$ on both sides, and otherwise we get $p_a$ on both sides.
In either case they're both equal, and since $p$ was arbitrary this means that the prime factorisation of $\operatorname{lcm}(a, \gcd(b, c))$ is the same as the prime factorisation of $\gcd(\operatorname{lcm}(a, b), \operatorname{lcm}(a, c))$, so the two numbers must be equal.
Here's a somewhat different approach. First, similar to what you did, the "if" part means each prime factor of $m$ is congruent to $1 \pmod{4}$. As shown in the answer to Sum of two squares and prime factorizations, Fermat's theorem on the sum of squares states each prime factor $p_i$ of $m$ can be written as the sum of squares. Also, for any $c, d, e, f \in \mathbb{R}$,
$$(c^2 + d^2)(e^2 + f^2) = (ce \pm df)^2 + (cf \mp de)^2 \tag{1}\label{eq1A}$$
shows whenever $2$ numbers can be written as a sum of squares, their product can be as well, in $2$ different ways. Using \eqref{eq1A} repeatedly with the previous result (starting at $1$) and for each $p_i \mid m$ means the final product, i.e., $m$, can be written as a sum of squares.
Regarding proving you can choose an $a$ and $b$ where $\gcd(a, b)$, the answer to Any product of primes in the form of 4n+1 is the sum of 2 relatively prime squares shows this, paraphrased below.
As shown in \eqref{eq1A}, the product of the $2$ sums of squares can be expressed in $2$ ways. Have $c^2 + d^2$, with $\gcd(c, d) = 1$, be a product of $1$ or more primes of the form $4n + 1$, and $e^2 + f^2$ be a prime of that form to be multiplied. Consider if the first form in \eqref{eq1A}, i.e., $(ce + df)^2 + (cf - de)^2$, is not valid, i.e., there's a prime $q$ which divides each term. This means
$$q \mid (ce + df)e + (cf - de)f = c(e^2 + f^2) \tag{2}\label{eq2A}$$
$$q \mid (ce + df)f - (cf - de)e = d(e^2 + f^2) \tag{3}\label{eq3A}$$
Since $q$ doesn't divide $c$ and $d$, then $q \mid e^2 + f^2 \implies q = e^2 + f^2$. If both solution types in \eqref{eq1A} are not valid, then $e^2 + f^2$ divides $ce - df$ as well as $ce + df$, and hence divides $2ce$ and $2df$. Since $e^2 + f^2$ doesn't divide $2e$ or $2f$, it must divide both $c$ and $d$, contrary to the hypothesis, meaning at least one of the $2$ forms must be valid. Thus, use the valid form, and repeat this procedure for each prime that is multiplied, to eventually get $m$.
For the "only if" part, similar to the answer to If $a \in \Bbb Z$ is the sum of two squares then $a$ can't be written in which of the following forms?, suppose there's a prime $p \equiv 3 \pmod{4}$ with $p \mid m$. If $p \mid a$, then $p \mid b$, and vice versa, but since $\gcd(a, b) = 1$, then $p$ can't divide either $a$ or $b$. Thus, $a$ has a multiplicative inverse, call it $a'$, modulo $p$. Let $r = \frac{p-1}{2}$ and note $r$ is odd. Also using Fermat's little theorem, this gives (note the argument below is basically equivalent to showing $-1$ is not a quadratic residue modulo $p$ if $p \equiv 3 \pmod{4}$)
$$\begin{equation}\begin{aligned}
a^2 + b^2 & \equiv 0 \pmod{p} \\
a^2(a')^2 + b^2(a')^2 & \equiv 0 \pmod{p} \\
1 + (ba')^2 & \equiv 0 \pmod{p} \\
(ba')^2 & \equiv -1 \pmod{p} \\
\left((ba')^2\right)^{r} & \equiv (-1)^r \pmod{p} \\
(ba')^{p-1} & \equiv -1 \pmod{p} \\
1 & \equiv -1 \pmod{p}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
This, of course, is not possible, meaning the original assumption must be false. This confirms all prime factors of $m$ must be congruent to $1 \pmod{4}$.
Best Answer
If we are several persons, the GREATEST height that we can all reach is the MINIMUM of our heights (the smallest person amongst us is the problem), the lowest door that we can all pass through is the MAXIMUM of our heights (the tallest person is the problem).