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?



Relation between commutator length and stable commutator length in free groups
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


