Let $$a, b, c \in \{1, 2, 3, 4\}$$. If the probability, that $$ax^2 + 2\sqrt{2}bx + c > 0$$ for all $$x \in \mathbb{R}$$, is $$\frac{m}{n}$$, $$\gcd(m, n) = 1$$, then $$m + n$$ is equal to __________.

The quadratic is $$f(x)= ax^{2}+2\sqrt{2}\,bx+c$$ with $$a,b,c \in\{1,2,3,4\}$$.

For $$f(x) \gt 0$$ for all real $$x$$, two conditions must hold:
1. Leading coefficient $$a \gt 0$$ (already true because $$a \ge 1$$).
2. Discriminant $$\Delta$$ is negative.

Compute the discriminant:
$$\Delta =(2\sqrt{2}b)^{2}-4ac = 8b^{2}-4ac =4\bigl(2b^{2}-ac\bigr).$$
Hence $$\Delta \lt 0 \Longleftrightarrow 2b^{2}-ac \lt 0 \Longleftrightarrow 2b^{2} \lt ac.$$

Total number of ordered triples $$(a,b,c)$$ is $$4^{3}=64.$$ We now count the favourable triples that satisfy $$2b^{2} \lt ac$$.

Condition becomes $$2b^{2} \lt c.$$
• $$b=1$$: $$2\lt c\Rightarrow c=3,4\;(2\text{ choices})$$
• $$b=2,3,4$$: $$2b^{2}\ge 8 \gt 4$$, no solutions.
Favourable triples = 2.

Condition becomes $$b^{2} \lt c.$$
• $$b=1$$: $$1\lt c\Rightarrow c=2,3,4\;(3\text{ choices})$$
• $$b=2,3,4$$ give $$b^{2}\ge 4$$, but $$c\le 4$$; no solutions.
Favourable triples = 3.

Condition becomes $$2b^{2} \lt 3c.$$ Check each $$c$$:
• $$c=1$$: need $$2b^{2}\lt 3 \Rightarrow b=1\;(1\text{ choice})$$
• $$c=2$$: need $$2b^{2}\lt 6 \Rightarrow b=1\;(1)$$
• $$c=3$$: need $$2b^{2}\lt 9 \Rightarrow b=1,2\;(2)$$
• $$c=4$$: need $$2b^{2}\lt 12 \Rightarrow b=1,2\;(2)$$
Favourable triples = $$1+1+2+2 = 6.$$\

Condition becomes $$b^{2} \lt 2c.$$ Check each $$c$$:
• $$c=1$$: $$b^{2}\lt 2 \Rightarrow b=1\;(1)$$
• $$c=2$$: $$b^{2}\lt 4 \Rightarrow b=1\;(1)$$
• $$c=3$$: $$b^{2}\lt 6 \Rightarrow b=1,2\;(2)$$
• $$c=4$$: $$b^{2}\lt 8 \Rightarrow b=1,2\;(2)$$
Favourable triples = $$1+1+2+2 = 6.$$\

Total favourable triples = $$2+3+6+6 = 17.$$

Therefore, required probability $$= \dfrac{17}{64}.$$ Here $$\gcd(17,64)=1$$, so $$m=17,\; n=64.$$ Finally, $$m+n = 17+64 = 81.$$

Answer: 81