Why does the wavelength of a particle go down with higher velocity? In mechanical waves, the wavelength increases with a higher velocity.
λ = v / f.
And the wavelength in particles lower with a higher velocity.
λ = h / mv.
 A: This is deceptive:
$λ = v / f$  for waves  that transfer energy with a fixed velocity v in a medium. This is a constant for the medium for all waves (acoustic, even light which is in vacuum with a constant  c velocity)

Suggested by De Broglie in about 1923, the path to the wavelength expression for a particle is by analogy to the momentum of a photon.

The appropriate  formula for particles in analogy with the photon is 

the de Broglie hypothesis which can be derived from the postulates of quantum mechanics. . The velocity represented by the momentum , $p$, is variable dependent on the condition of the particle.  The wave nature of the particle  is a probability wave, not a mass or energy wave. 
A single particle does not register as a wave, it is always a point in (x,y,z,t). An accumulation of particles with the same boundary conditions will show wave behavior in the probability distribution. See this answer to see how one measures a probability distribution for a particle.
A: The higher the frequency the smaller the wavelength. The higher velocity equates to higher acceleration creating higher energy (higher frequency)photons. Smaller wavelength means higher energy.
A: The de Broglie wavelength is the central wavelength of a superposition of waves forming a wave packet. The velocity in that equation is the envelope velocity of all of the combined waves (the group velocity), and differs from the individual wave velocities.
https://en.wikipedia.org/wiki/Group_velocity 
A: The short answer is "because": the dispersion relations dictate it. I'll illustrate @anna's answer in elementary mathematese, contrasting water gravity waves, in the deep ocean, to a QM particle in Schroedinger's equation.
The dispersion relation is the functional dependence of the angular frequency on the wavenumber, $\omega(k)$. The group velocity of a wavepacket is 
$$v=\frac{\partial \omega}{\partial k }. $$
The deep water wave dispersion relation is 
$$
\omega=\sqrt{g k},
$$ 
easily derived in books, but also from dimensional analysis, having been reassured, from generic thinking (if you trust yourself) that the only relevant constant involved is the acceleration of gravity g, and nothing else. Given that, deep water waves necessarily travel with
$$
v=\frac{\omega}{2k}\propto \sqrt{\lambda} \propto 1/\sqrt{\omega},
$$
familiar from handicapping sailboats, or from the fact short frequency waves from a boat in a lake reach you last as you are sitting on the shore! [Your expression is peculiar]. (This dispersion relation is also the origin of the universal Kelvin wake angle of all swimming/sailing objects, independent of velocity!)
Linear dispersion relations ($\omega  = a k$), for constant a, of course, give you constant speeds $v=a$--possibly that's your first equation, a possible misconception anna identified.
The dispersion relation of QM de Broglie waves from Schroedinger's equation
is radically different, 
$$
\omega=\frac{\hbar k^2}{2m}, 
$$ 
again dimensionally inevitable, once an Austrian gentleman reassures you these are the only constants involved. [Let's not really go there.]
The resulting speed is then 
$$
v=\hbar k/m=\frac{h}{m} \frac {1}{\lambda} ~ . 
$$ 
This is the mother of all resolution experiments, so increasing the speed and thus energy of your probing projectiles involves shorter wavelengths proving shorter distances.
