Question Number 148226 by mathdanisur last updated on 26/Jul/21
Answered by mitica last updated on 26/Jul/21
$$\Sigma\frac{\mathrm{1}}{{x}\left({px}+\mathrm{1}\right)}=\Sigma\frac{\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} }{{p}+\frac{\mathrm{1}}{{x}}}\geqslant \\ $$$$\frac{\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}\right)^{\mathrm{2}} }{{p}+{q}+{r}+\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}}=\frac{\mathrm{9}}{{p}+{q}+{r}+\mathrm{3}} \\ $$
Commented by mathdanisur last updated on 26/Jul/21
$${Thank}\:{you}\:{Sir} \\ $$