To prove the inequality sin$\pi /3-2x$cos$\pi /3-2x$ ≥ -√3/4, we can use the double angle formula for sine and cosine:

sin$2\theta$ = 2sin$\theta$cos$\theta$ cos$2\theta$ = cos^2$\theta$ - sin^2$\theta$

Let's first substitute θ = π/6 - x into the double angle formulas:

sin$\pi /3-2x$ = 2sin$\pi /6-x$cos$\pi /6-x$ cos$\pi /3-2x$ = cos^2$\pi /6-x$ - sin^2$\pi /6-x$

Now, substitute these expressions back into the original inequality:

2sin$\pi /6-x$cos$\pi /6-x$$co{s}^2(\pi /6-x)-si{n}^2(\pi /6-x)$ ≥ -√3/4

Expand the expression and simplify:

2sin$\pi /6-x$cos$\pi /6-x$cos^2$\pi /6-x$ - 2sin^3$\pi /6-x$ ≥ -√3/4

Since sin$\pi /6$ = 1/2 and cos$\pi /6$ = √3/2, the inequality simplifies to:

2$1/2-si{n}^3(\pi /6-x)$√3/2 - 2sin^3$\pi /6-x$ ≥ -√3/4

√3 - 3sin^3$\pi /6-x$√3 - 2sin^3$\pi /6-x$ ≥ -√3/4

√3 - 5sin^3$\pi /6-x$√3 ≥ -√3/4

1 - 5sin^3$\pi /6-x$ ≥ -1/4

5sin^3$\pi /6-x$ ≤ 5/4

sin^3$\pi /6-x$ ≤ 1/4

Since -1 ≤ sin$x$ ≤ 1, and sin$\pi /6-x$ is in the range of sin$x$, we have established that sin^3$\pi /6-x$ ≤ 1/4. Thus, the inequality sin$\pi /3-2x$cos$\pi /3-2x$ ≥ -√3/4 holds true.