Derive chernoff bound for $\mathrm{erfc}(x)$
I don't know why you are approaching the problem the way you are by going to a two-dimensional approach.

The Chernoff bound starts with the observation that for any random variable
$X$ and any $\lambda &gt; 0$,
$$P\{X \geq a\} \leq E[\exp(\lambda(X-a))]\tag{1}$$
because $\exp(\lambda(x-a))\geq  \mathbf 1_{\{x\colon x \geq a\}}$.
Consequently, 
$$P\{X \geq a\} \leq \min_{\lambda &gt; 0}E[\exp(\lambda(X-a))].\tag{2}$$
For a standard Gaussian random variable, $E[\exp(\lambda(X-a))]$
can be found easily by using a method called "completing the square"
(in the exponent which as terms like $-\frac{1}{2}x^2+\lambda(x-a)$ in
it) and noting that the resulting integral is the integral
of a Gaussian density times a constant. Alternatively, you can
use the moment-generating function and read off the answer.
After minimization with respect to $\lambda$, the result is
$$P\{X \geq a\} = Q(x) \leq e^{-\frac{x^2}{2}}.\tag{3}$$
The two-dimensional approach leads to the slightly better result
$$Q(x) \leq \frac 12e^{-\frac{x^2}{2}}.\tag{4}$$
Note that you have proved already that for $x &gt; 0$,
$$P\{X_1&gt;x, X_2&gt;x\} = Q^2(x) \leq \frac 14 e^{-{x^2}}$$
from which $(4)$ follows immediately, and $(4)$ is not the
Chernoff bound on $Q(x)$.