The inequation is:
$$g = c_1 (x - l_1(x - 1)) + c_2 (x + l_2(x - 1)) > 1$$
The variable term is unsurprisingly x, the term that must be resolved. The other variables are all constants. The "preconditions" are:
$$l_1,l_2 \in \mathbb{N}_{>0}$$ $$x \in \mathbb{R_{\geq 0}}$$ $$c_1,c_2 \in \mathbb{R}_{[0,1]}$$ $$c_1 + c_2 \leq 1$$
In other words, $l_1$ and $l_2$ are multipliers; they magnify the effect of each summed term. $c_1$ and $c_2$ represent percentual weights over a common total, and usually $c_2 = 1 - c_1$. Concretely, we can "provisionaly" assume that:
$$c_1 + c_2 = 1$$
So, let's go:
$$c_1(x - l_1x + l_1) + c_2(x + l_2x - l_2) > 1 $$ $$c_1(x(1 - l_1) + l_1) + c_2(x(1 + l_2) - l_2) > 1$$ $$x(c_1(1-l_1) + c_2(1+l_2)) + c_1l_1 - c_2l_2 > 1$$ $$x > \frac{1 - (c_1l_1 - c_2l_2)}{c_1(1-l_1) + c_2(1+l_2)}$$ $$x > \frac{1-(c_1l_1 - c_2l_2)}{(c_1+c_2)-(c_1l_1 - c_2l_2)}$$
Which means than multipliers doesn't matter. If $c_1 + c_2 = 1$, then $x > 1$ if you want that $g > 1$.
But anyway, despite $c_1,c_2$ summing $1$ together or not, something is wrong with the inequation:
$$(l_1,l_2) = (2,1)$$ $$(c_1,c_2) = (0.7,0.3)$$ $$x = 1.1 > 1$$ $$0.7 (1.1 - 2(1.1 - 1)) + 0.3 (1.1 + 1(1.1 - 1)) = 0.99 < 1$$
I've spent tons of hours already seeking for a flaw but I found none. I know that the denominator must be $!= 0$ for the inequation to apply, but that's uncommon, and anyway, it doesn't happen for this specific example. Besides that, I'm blind right know.
It's not that the denominator must be $\neq 0$, for your manipulation to be valid it must be $>0$. From $$x(c_1(1-l_1)+c_2(1+l_2))+c_1l_1-c_2l_2>1$$ you get $$x>\frac{1-(c_1l_1-c_2l_2)}{c_1(1-l_1)+c_2(1+l_2)}$$ if the denominator is positive, but $$x<\frac{1-(c_1l_1-c_2l_2)}{c_1(1-l_1)+c_2(1+l_2)}$$ if it is negative (as it is for your example).
(Multiplying or dividing both sides of an inequality by the same amount reverses the inequality if that amount is negative.)