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The measures of skewness and kurtosis respectively are
$b_1=\frac{m_3^2}{m_2^3}$(skewness)
and
$b_2=\frac{m_4}{m_2^2}$(Kurtosis)
where $m_r$ is the central moment of $rth$ order. That is $m_r = \frac{\sum_{i=1}^{n}(x_i-\bar{x})^r}{n}$
$\bar{x}= \frac{\sum_{i=1}^{n}x_i}{n}$
To Prove:
$b_2 -1\geq b_1$
I was trying to use Cauchy-schwarz inequality by letting $a_i=(x_i -\bar{x})^2$ ; $b_i =1$
and get that $b_2 \geq 1$
But it is getting very much complicated after that when working with $b_1$
Any idea would be good for me to make the proof simpler.
Thank you.
EDIT: I look for an answer or hint associated with Cauchy Schwarz or AM-GM inequality. The answers here Prove that Kurtosis is at least one more than the square of the skewness uses matrix but I don't know matrix that much. I am very new to statistics.

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    It does not affect your question, but your $b_1$ is essentially the square of skewness rather than just skewness. – Henry Mar 15 '24 at 10:25
  • Related: https://stats.stackexchange.com/questions/512568/prove-that-kurtosis-is-at-least-one-more-than-the-square-of-the-skewness – Henry Mar 15 '24 at 10:28

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