TITLE: Relation between commutator length and stable commutator length in free groups
QUESTION [6 upvotes]: In Bardakov, Algebra and Logic, Vol. 39, No. 4, 2000 I have found the following (page 225, see https://link.springer.com/article/10.1007/BF02681648)

We pronounce tile validity of the following:
Conjecture. For every element z in the derived subgroup of a free non-Abelian group F and for any natural m,
  $$
\mathrm{cl}(z^m) \geq (m+1/2)\mathrm{cl}(z)
$$

Where cl denotes the commutator length of an element (ie. the minimal number to express it as a product of commutators). 
This inequality is not true, and $$z = [a, b]$$ may be a counterexample. However, I belive that there may be a typo, so it should rather be
$$
\mathrm{cl}(z^m) \geq (m+1)/2 \cdot \mathrm{cl}(z)
$$
Unvortunatelly, I could not find it in any other paper/book (including Calegari's "scl"). And the proof in Bardakov is unclear to me. 
Do you know any paper, with a proof of the above inequality? Or maybe some counterexample? Or maybe has anybody have any clue why Bardakov did not prove this inequality?

REPLY [2 votes]: The misconception is due to bad translation. The original text was in Russian and can be found here: http://www.mathnet.ru/links/f40b0cf29e7b19a9b2ab1a95ef70baba/al284.pdf. 
It reads: "Можно высказать предположение о том" which means "we can phrase a guess". 
I e-mailed Valeiry Bardakov about that conjecture. And it occured to be false. In fact there exists a sequence $z_n$ of elements of a free group such that $cl(z_n) \to \infty$, but $cl(z_n^2) < const$. It can be seen using the result by O.Kharlampovich and A.Myasnikov [1,Section 6 and 7].
According to Theorem 3 of [1], there exist solutions to 
$$x_1^2 x_2^2 x_3^2 x_4^2 = 1 \quad (*)$$ 
such that $cl(x_1x_2x_3x_4)$ is arbitrarily big. But we can observe, that $(x_1x_2x_3x_4)^2 = x_1x_2x_3x_4 x_1x_2x_3x_4$ can be expressed as $x_1^2 x_2^2 x_3^2 x_4^2$ times a bounded number of commutators. And now we see that (*) implies that $cl((x_1x_2x_3x_4)^2)$ is bounded by a fixed constant, that we can easily calculate.  
[1]O.Kharlampovich and A.Myasnikov, "Implicit Function Theorem over Free
Groups and Genus Problem", in: Knots, braids, and mapping class
groups—papers dedicated to Joan S.
Birman (New York, 1998), Amer. Math. Soc., Providence, RI, 2001, 77–83