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José Hdz. Stgo.
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In Encyclopedia of Mathematicsthe Encyclopedia of Mathematics there is an inequality for Levy metric ($d_L$): $$d_L(E,F) \leq \{\beta_r(F)\}^{r/(r+1)},$$ where E$E$ is a a distribution that is degenerate at zero, $\beta_r(F)$, $r>0$ is an absolute moment of the distribution $F$ (http://www.encyclopediaofmath.org/index.php/L%C3%A9vy_metric). I cannot prove this inequality. I can get only $$d_L(E,F) \leq \{\beta_r(F)\}^{1/(r+1)}.$$ Do you have any ideas how to get $r/(r+1)$?

In Encyclopedia of Mathematics there is an inequality for Levy metric ($d_L$): $$d_L(E,F) \leq \{\beta_r(F)\}^{r/(r+1)},$$ where E is a a distribution that is degenerate at zero, $\beta_r(F)$, $r>0$ is an absolute moment of the distribution $F$ (http://www.encyclopediaofmath.org/index.php/L%C3%A9vy_metric). I cannot prove this inequality. I can get only $$d_L(E,F) \leq \{\beta_r(F)\}^{1/(r+1)}.$$ Do you have any ideas how to get $r/(r+1)$?

In the Encyclopedia of Mathematics there is an inequality for Levy metric ($d_L$): $$d_L(E,F) \leq \{\beta_r(F)\}^{r/(r+1)},$$ where $E$ is a a distribution that is degenerate at zero, $\beta_r(F)$, $r>0$ is an absolute moment of the distribution $F$ (http://www.encyclopediaofmath.org/index.php/L%C3%A9vy_metric). I cannot prove this inequality. I can get only $$d_L(E,F) \leq \{\beta_r(F)\}^{1/(r+1)}.$$ Do you have any ideas how to get $r/(r+1)$?

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Estimate for Levy metric

In Encyclopedia of Mathematics there is an inequality for Levy metric ($d_L$): $$d_L(E,F) \leq \{\beta_r(F)\}^{r/(r+1)},$$ where E is a a distribution that is degenerate at zero, $\beta_r(F)$, $r>0$ is an absolute moment of the distribution $F$ (http://www.encyclopediaofmath.org/index.php/L%C3%A9vy_metric). I cannot prove this inequality. I can get only $$d_L(E,F) \leq \{\beta_r(F)\}^{1/(r+1)}.$$ Do you have any ideas how to get $r/(r+1)$?