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

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