The given inequality is
$$\sqrt{m^2 + 1} + \sqrt{(3-m)^2 + 1} \leq \sqrt{\frac{2(4m^2 - 12m -15}{(m-3)m}} \ \text{where,} \ m \in [0, 3]$$
I took the following functions
$$f(m) = \sqrt{m^2 + 1} + \sqrt{(3-m)^2 + 1}$$
$$\text{and,}\ g(m) = \frac{2(4m^2 - 12m -15}{(m-3)m}$$
The first derivative of $f(m)$ is
$$f^{\prime}(m) = \frac{m}{\sqrt{m^2 + 1}} + \frac{(m - 3)}{\sqrt{(3 - m)^2 + 1}}$$
Then I solved for $f^{\prime}(m) \gt 0$ since it would tell you what the point of maxima or minima is for the given interval via the first derivative test.
$$f^{\prime}(m) \gt 0$$
$$\implies \frac{m}{\sqrt{m^2 + 1}} + \frac{(m - 3)}{\sqrt{(3 - m)^2 + 1}} \gt 0$$
$$\implies m\sqrt{(3-m)^2+1} + (m-3)\sqrt{m^2+1} \gt 0$$
$$\implies m\sqrt{(3-m)^2+1} \gt -(m-3)\sqrt{m^2+1}$$
$$\implies m^2 ((3-m)^2+1) \gt (m-3)^2 (m^2+1)$$
$$\implies m^4 - 6m^3 + 10m^2 > m^4 - 6m^3 + 10m^2 - 6m + 9$$
$$\implies m \gt \frac{3}{2}$$
Thus, $m = 1.5$ is a point of local minimum. And $f(0) = \sqrt{10} + 1\approx 4.162\dots$, $f(3) = \sqrt{10} + 1\approx 4.162\dots$, $f(1.5) = 2\sqrt{3.25} \approx 3.605\dots$.
$$\therefore f(m) \in [3.605\dots, 4.162\dots]\ \forall \ m \in [0, 3]$$
And doing the same for $g(m)$, we will get
$$\begin{align}
g^{\prime}(m) &= \frac{2m(m-3)(8m-12) - 2(2m-3)(4m^2 -12m -15)}{(m-3)^2 m^2}\\
&=\frac{(2m-3)\cdot(8m^2 - 24m - 8m^2 + 24m + 30)}{(m-3)^2 m^2}\\
&=\frac{30(2m-3)}{(m-3)^2 m^2}
\end{align}$$
Then solve for $g^{\prime}(m) \gt 0$ to get $m \gt 1.5$, which will be a point of local minimum. If I define a function
$$h(m) = \sqrt{g(m)} - 1$$
Then we will have $h(1.5) = 8/\sqrt{3} - 1 \approx 3.618\dots$, $h(0)$ and $h(3)$ approach positive infinity.
$$\therefore h(m) \in [3.618\dots, \infty] \ \forall \ m \in [0,3]$$
Since $f(m)$ and $h(m)$ never intersect, we can conclude that $f(m) \leq h(m) \ \forall \ m \in [0,3]$.