Question Number 193238 by lmcp1203 last updated on 08/Jun/23
$${s}={a}+{b}+{c}+{d}+….. \\ $$$${number}\:{terms}\::{n} \\ $$$$\left\{{a};{b};{c};{d}…..\right\}>\mathrm{0} \\ $$$$\left.{then}\:{E}={s}/\left({s}−{a}\right)+{s}/\left({s}−{b}\right)+{s}/{s}−{c}\right)+…. \\ $$$$\left.{a}\left.\right)\:{E}>={n}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:{b}\right){E}>={n}^{\mathrm{2}} /\left({n}−\mathrm{1}\right) \\ $$$$\left.{c}\left.\right)\:{E}>={n}/\left({n}+\mathrm{1}\right)\:\:\:\:\:\:{d}\right)\:{E}>={n}^{\mathrm{2}} /\left({n}+\mathrm{1}\right) \\ $$$$\left.{e}\right)\:{E}>={n}^{\mathrm{2}} −\mathrm{1} \\ $$$$ \\ $$$$ \\ $$
Answered by MM42 last updated on 08/Jun/23
$${E}\geqslant{n}\sqrt[{{n}}]{\frac{{s}^{{n}} }{\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)…}}\:=\frac{{ns}}{\:\sqrt[{{n}}]{\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)…}}\: \\ $$$$\geqslant\frac{{n}^{\mathrm{2}} {s}}{\left({s}−{a}\right)+\left({s}−{b}\right)+\left({s}−{c}\right)+…}=\frac{{n}^{\mathrm{2}} {s}}{{ns}−{s}}=\frac{{n}^{\mathrm{2}} }{{n}−\mathrm{1}}\:\:\checkmark\:\left({b}\right) \\ $$$$ \\ $$
Commented by lmcp1203 last updated on 08/Jun/23
$${thank}\:{you} \\ $$