Is it true that high-bypass engines are inherently inefficient at higher speeds?
Power consumption=Kinetic energy=$1/2mv^2$.
  Thrust=momentum change=$mv$.


This is only true if the engine is perfectly efficient internally and the air is still relative to the aircraft.

If the air is moving relative to the aircraft (and the engine is still perfectly efficient internally) then.

Power consumption=Kinetic energy=$\frac{1}{2}mv_e^2 - \frac{1}{2}mv_a^2$ = $\frac{1} {2}m((v_a + v_\Delta)^2 - v_a^2) = \frac{1} {2}m(2v_av_\Delta + v_\Delta^2) $ 
Thrust=momentum change=$m(v_e-v_a) = mv_\Delta$

Where $v_a$ is the ambient velocity (relative to the aircraft), $v_e$ is the exhaust velocity (relative to the aircraft) and $v_\Delta$ is the difference between the ambient velocity and the exhaust velocity ($v_e - v_a$).

If the exhaust velocity (relative to the aircraft) is smaller than the ambient velocity (relative to the aircraft) then you have negative thrust (aka drag).

If the exhaust velocity (relative to the aircraft) is much larger than the ambient velocity (relative to the aircraft) then most of the power is wasted (the $v_\Delta^2$ term is larger than the $2v_av_\Delta$ term)

If the exhaust velocity is only slightly higher than the ambient velocity then most of the power goes into useful thrust (the $2v_av_\Delta$ term is larger than the $v_\Delta^2$ term)

What this means if that if the engine were perfectly efficient internally then the advantages of a higher bypass ratio would drop off with speed, but there would still be a slight advantage to the engine with higher bypass ratio.

But that is a big if, in particular we assume that all the energy in the intake stream is captured and returned to the exhaust stream. 



Lets consider what happens if we instead assume that the energy in the intake stream is lost. 

Our power consumption equation becomes.

Power consumption=Kinetic energy=$\frac{1}{2}mv_e^2$ = $\frac{1} {2}m(v_a + v_\Delta)^2) = \frac{1} {2}m(v_a^2 + 2v_av_\Delta + v_\Delta^2) $ 

For a given thrust $m$ is proportional to $\frac{1}{v_\Delta}$ so our goal is to minimise. $\frac{1}{v_\Delta}(v_a^2 + 2v_av_\Delta + v_\Delta^2)$ = $\frac{v_a^2}{v_\Delta} + 2v_a + v_\Delta$ differentiating with respect to v_\Delta and setting to 0 gives $-\frac{v_a^2}{v_\Delta^2} + 1$ which (given that v_\Delta must be positive) gives $v_a = v_\Delta$



Having considered an optimistic and a pessimitic case we can conclude that the optimal exhaust velocity for efficency is greater than the ambient velocity (otherwise you would not produce thurst) but probably less than twice the ambient velocity.