Explaining this inequality

Question

In a proof I was working on today, I assumed this equation was true which lead to devastating results

$$ \sqrt{\bar{x^2}} =\bar{\lvert x\rvert} $$

For instance, given the data 0 and 2, the left hand side results in $\sqrt{2}$ whereas the right hand side results in $1$. I know that if I put the average sign on the left above the root they would be exactly the same (because that's what absolute is defined as) but what is wrong with bringing the average sign into the root?

Answer 1

If I understand well your notation, the left-hand side is the quadratic mean, while the right-hand side is the arithmetic mean. It is well known is less than the former.

More generally, if we denote by $\,m_p(x)=\biggl(\dfrac{\sum_{i=1}^n x_i^p}n\biggr)^{\!\tfrac1p}\,$ the $p$-power mean, it can be proved that $$p<q \Rightarrow m_p(x)\le m_q(x)$$ and we have equality if and only if all $x_i$ are equal.

Note that $m_1$ is the arithmetic mean and $m_{-1}$ is the harmonic mean.

Answer 2

That "bar" must be inside square sign too. This can be explained by the famous "Cauchy-Schwarz inequality". But in simple terms, LHS is "Root mean square value" while RHS is just "mean value"