Move all expressions to one side of the inequality:
f(x)=12^x-5-6^(1-x)>0
Taking the derivative of the left side with respect to x we have:
f'(x)=log(12)12^x+log(6)6^(1-x)
This expression is positive for all x in RR, therefore f(x) is strictly increasing at all times. This leaves two possibilities:
1) The expression is always positive and therefore the inequality is satisfied for all x in RR
2) The expression has a single root and satisfies the inequality for any x greater than that root.
If we can find any value of f(x) that is negative then the second possibility is true. We can see this is true for f(0):
f(0)=12^0-5-6^(1-0)=-10
If we can solve f(x)=0 then we solve this problem. There is no obvious way I can see to analytically solve this expression so I would use a numerical method. if you use the Newton-Raphson method (https://en.wikipedia.org/wiki/Newton%27s_method) with an initial guess of x_0=0 then the series:
x_(n+1)=x_n-f(x_n)/(f'(x_n))
Will converge on the root. We find that this converges after a few iterations:
x=0.756214 - 6 s.f.
Therefore we have the solution:
x>0.756214 knowing this is a numerical approximation to the solution and only accurate to 6 s.f..