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If $ P(z)=a_n\prod _{v=1}^{n}(z-z_v) $ P (z) = a n ∏ v = 1 n (z − z v) is a polynomial of degree n having all its zeros in $ |z|\le k, k\ge 1 $ | z | ≤ k , k ≥ 1 then Aziz [Inequalities for the derivative of a polynomial. Proc Am Math Soc. 1983;89(2):259–266] proved that \[ \max_{|z|=1}|P'(z)|\ge \frac{2}{1+k^n}\sum_{v=1}^{n}\frac{k}{k+|z_v|}\max_{|z|=1}|P(z)|. \] max | z | = 1 | P ′ (z) | ≥ 2 1 + k n ∑ v = 1 n k k + | z v | max | z | = 1 | P (z) |. Recently, Kumar [On the inequalities concerning polynomials. Complex Anal Oper Theory. 2020;14(6):1–11 (Article ID 65)] established a generalization of this inequality and proved under the same hypothesis for a polynomial $ P(z)=a_0+a_1z+a_2z^2+\cdots +a_nz^n=a_n\prod _{v=1}^{n}(z-z_v) $ P (z) = a 0 + a 1 z + a 2 z 2 + ⋯ + a n z n = a n ∏ v = 1 n (z − z v) , that $$\begin{align*} & \max_{|z|=1}|P'(z)| \\ & \ge \left(\frac{2}{1+k^n}+\frac{(|a_n|k^n-|a_0|)(k-1)}{(1+k^n)(|a_n|k^n+k|a_0|)}\right)\sum_{v=1}^{n}\frac{k}{k+|z_v|}\max_{|z|=1}|P(z)|. \end{align*} $$ max | z | = 1 | P ′ (z) | ≥ (2 1 + k n + (| a n | k n − | a 0 |) (k − 1) (1 + k n) (| a n | k n + k | a 0 |)) ∑ v = 1 n k k + | z v | max | z | = 1 | P (z) |. In this paper, we sharpen the above inequalities and further extend the obtained results to the polar derivative of a polynomial. As a consequence, our results also sharpens considerably some results of Dewan and Upadhye [Inequalities for the polar derivative of a polynomial. J Ineq Pure Appl Math. 2008;9:1–9 (Article ID 119)]. [ABSTRACT FROM AUTHOR]