Problem
Let $R$ and $s$ be respectively the range and standard deviation , respectively, of a set of $n$ values. Then prove that $\dfrac{R^2}{2n}\leq s^2\leq \dfrac{R^2}{4}$.
I have tried using the definition of range and standard deviation but neither of the inequality can be proved. Can anyone help me in proving it?
Without loss of generality (by translation), assume that the data is such that $0\leq x_i\leq R$, with the smallest data $x_{\min}=0$, and the largest data is $x_{\max}=R$. Let $\mu$ be the mean. Then $$\sum_i x_i^2 = \sum_i x_ix_i \leq \sum_i Rx_i= nR\mu$$ Also note that $\mu\leq R$. Therefore, $$ns^2= \sum_i (x_i-\mu)^2 = \sum_i x_i^2 -n\mu^2 \leq nR\mu-n\mu^2 =n\mu(R-\mu)\leq nR^2/4.$$ So, $s^2\leq R^2/4$.
For the lower bound, note that $$2ns^2= 2\sum_i (x_i-\mu)^2 \geq 2[(x_{\max}-\mu)^2+ (x_{\min}-\mu)^2]=2[(R-\mu)^2+\mu^2]\geq R^2,$$ where the last inequality follows from arithmetic-quadratic mean inequality, which is $$\frac{x_1^2+\ldots+x_n^2}{n} \geq \left(\frac{x_1+\ldots+x_n}{n}\right)^2$$ applied to $(R-\mu)$ and $\mu$. Therefore, $s^2\geq R^2/2n$, and so this completes the proof.