x_n>ℓ → x_n - ℓ>0 → [x_(n+1)]² - ℓ²>0 → [x_(n+1)]²>ℓ²

x_n>ℓ → x_n>0 → x_(n+1)>0

Since both x_(n+1) and ℓ are positive, [x_(n+1)]²>ℓ² → x_(n+1)>ℓ

[x_(n+1)]² - ℓ² = x_n - ℓ

(x_(n+1) + ℓ)(x_(n+1) - ℓ) = x_n - ℓ

Since x_(n+1)>ℓ → x_(n+1) + ℓ > 1,

x_n - ℓ > x_(n+1) - ℓ

x_n > x_(n+1)

∴x_n > x_(n+1) > ℓ

I've not learnt JC math, so idk if there's a shorter way