If F(z) is a polynomial of degree n having all zeros in |z|≤k,k>0\documentclass[12pt]{minimal}
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\begin{document}$$|z|\le k,~k>0$$\end{document} and f(z) is a polynomial of degree m≤n\documentclass[12pt]{minimal}
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\begin{document}$$m\le n$$\end{document} such that |f(z)|≤|F(z)|\documentclass[12pt]{minimal}
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\begin{document}$$|f(z)|\le |F(z)|$$\end{document} for |z|=k\documentclass[12pt]{minimal}
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\begin{document}$$|z|=k$$\end{document}, then it was formulated by Rather and Gulzar (Adv Inequal Appl 2:16–30, 2013) that for every |δ|≤1,|β|≤1,R>r≥k\documentclass[12pt]{minimal}
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\begin{document}$$|\delta |\le 1, |\beta |\le 1,~R>r\ge k$$\end{document} and |z|≥1,\documentclass[12pt]{minimal}
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\begin{document}$$|z|\ge 1,$$\end{document}|B[foσ](z)+ψB[foρ](z)|≤|B[Foσ](z)+ψB[Foρ](z)|,\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} |B[fo\sigma ](z)+\psi B[fo\rho ](z)|\le |B[Fo\sigma ](z)+\psi B[Fo\rho ](z)|, \end{aligned}$$\end{document}where B is a Bn\documentclass[12pt]{minimal}
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\begin{document}$$B_{n}$$\end{document} operator, σ(z)=Rz,ρ(z)=rz\documentclass[12pt]{minimal}
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\begin{document}$$\sigma (z){=}Rz, \rho (z){=}rz$$\end{document} and ψ:=ψ(R,r,δ,β,k)=β{(R+kr+k)n-|δ|}-δ\documentclass[12pt]{minimal}
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\begin{document}$$\psi {:=}\psi (R,r,\delta ,\beta ,k) {=}\beta \bigg \{\bigg (\frac{R+k}{r+k}\bigg )^{n}{-}|\delta |\bigg \}{-}\delta $$\end{document}. The authors have assumed that B∈Bn\documentclass[12pt]{minimal}
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\begin{document}$$B\in B_{n}$$\end{document} is a linear operator which is not true in general. In this paper, besides discussing assumption of authors and their followers (see e.g, Rather et al. in Int J Math Arch 3(4):1533–1544, 2012), we present the correct proof of the above inequality. Moreover our result improves many prior results involving Bn\documentclass[12pt]{minimal}
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\begin{document}$$B_{n}$$\end{document} operators and a number of polynomial inequalities can also be deduced by a uniform procedure.