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Question Number 194037 by Subhi last updated on 26/Jun/23
Let a_1 ,a_2 ....a_n ∈R^+ , a_1 +a_2 +.....a_n =1 prove that: (a_1 /(2−a_1 ))+(a_2 /(2−a_2 )).......(a_n /(2−a_n ))≥(n/(2n−1))
$$ \\ $$$${Let}\:{a}_{\mathrm{1}} ,{a}_{\mathrm{2}} ….{a}_{{n}} \in{R}^{+} ,\:{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +…..{a}_{{n}} =\mathrm{1} \\ $$$${prove}\:{that}: \\ $$$$\frac{{a}_{\mathrm{1}} }{\mathrm{2}−{a}_{\mathrm{1}} }+\frac{{a}_{\mathrm{2}} }{\mathrm{2}−{a}_{\mathrm{2}} }…….\frac{{a}_{{n}} }{\mathrm{2}−{a}_{{n}} }\geqslant\frac{{n}}{\mathrm{2}{n}−\mathrm{1}} \\ $$
Answered by AST last updated on 26/Jun/23
WLOG,Let a_1 ≥a_2 ≥...≥a_n ⇒(1/(2−a_1 ))≥(1/(2−a_2 ))≥...≥(1/(2−a_n )) Chebyshev⇒Σ(a/(2−a_1 ))≥(1/n)(Σa_1 )(Σ(1/(2−a_1 )))=(1/n)Σ((1/(2−a_1 ))) ≥(1/n)((n^2 /(2n−Σa_1 )))=(n/(2n−1))
$${WLOG},{Let}\:{a}_{\mathrm{1}} \geqslant{a}_{\mathrm{2}} \geqslant…\geqslant{a}_{{n}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}−{a}_{\mathrm{1}} }\geqslant\frac{\mathrm{1}}{\mathrm{2}−{a}_{\mathrm{2}} }\geqslant…\geqslant\frac{\mathrm{1}}{\mathrm{2}−{a}_{{n}} } \\ $$$${Chebyshev}\Rightarrow\Sigma\frac{{a}}{\mathrm{2}−{a}_{\mathrm{1}} }\geqslant\frac{\mathrm{1}}{{n}}\left(\Sigma{a}_{\mathrm{1}} \right)\left(\Sigma\frac{\mathrm{1}}{\mathrm{2}−{a}_{\mathrm{1}} }\right)=\frac{\mathrm{1}}{{n}}\Sigma\left(\frac{\mathrm{1}}{\mathrm{2}−{a}_{\mathrm{1}} }\right) \\ $$$$\geqslant\frac{\mathrm{1}}{{n}}\left(\frac{{n}^{\mathrm{2}} }{\mathrm{2}{n}−\Sigma{a}_{\mathrm{1}} }\right)=\frac{{n}}{\mathrm{2}{n}−\mathrm{1}} \\ $$
Commented by Subhi last updated on 26/Jun/23
perfect
$${perfect}\: \\ $$
Answered by Subhi last updated on 26/Jun/23
Another solution (a_1 ^2 /(2a_1 −a_1 ^2 ))+(a_2 ^2 /(2a_2 −a_2 ^2 ))......(a_n ^2 /(2a_n −a_n ^2 ))≥(((a_1 +a_2 ...a_n )^2 )/(2(a_1 +a_2 .....a_n )−(a_1 ^2 +a_2 ^2 ...a_n ^2 ))) ≥ (1/(2−(((a_1 +a_2 ....a_n )^2 )/n)))=(1/(2−(1/n)))=(n/(2n−1))
$${Another}\:{solution} \\ $$$$\frac{{a}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{2}{a}_{\mathrm{1}} −{a}_{\mathrm{1}} ^{\mathrm{2}} }+\frac{{a}_{\mathrm{2}} ^{\mathrm{2}} }{\mathrm{2}{a}_{\mathrm{2}} −{a}_{\mathrm{2}} ^{\mathrm{2}} }……\frac{{a}_{{n}} ^{\mathrm{2}} }{\mathrm{2}{a}_{{n}} −{a}_{{n}} ^{\mathrm{2}} }\geqslant\frac{\left({a}_{\mathrm{1}} +{a}_{\mathrm{2}} …{a}_{{n}} \right)^{\mathrm{2}} }{\mathrm{2}\left({a}_{\mathrm{1}} +{a}_{\mathrm{2}} …..{a}_{{n}} \right)−\left({a}_{\mathrm{1}} ^{\mathrm{2}} +{a}_{\mathrm{2}} ^{\mathrm{2}} …{a}_{{n}} ^{\mathrm{2}} \right)} \\ $$$$\geqslant\:\frac{\mathrm{1}}{\mathrm{2}−\frac{\left({a}_{\mathrm{1}} +{a}_{\mathrm{2}} ….{a}_{{n}} \right)^{\mathrm{2}} }{{n}}}=\frac{\mathrm{1}}{\mathrm{2}−\frac{\mathrm{1}}{{n}}}=\frac{{n}}{\mathrm{2}{n}−\mathrm{1}} \\ $$
Answered by witcher3 last updated on 26/Jun/23
f(x)=(1/(2−x)),f′(x)=(1/((2−x)^2 ))>0 f′′=(2/((2−x)^3 ))≥0 f convex ⇒a_i f(a_i )≥f(Σa_i ^2 )...E Quadratic Mean ⇒(((Σa_i ^2 )/n))^(1/2) ≥((Σa_i )/n) Σa_i ^2 ≥(1/n)...f increase function ⇒f(Σa_i ^2 )≥f((1/n)) E⇒Σ_(i=1) ^n (a_i /(2−a_i ))≥f((1/n))=(n/(2n−1))
$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}−\mathrm{x}},\mathrm{f}'\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{x}\right)^{\mathrm{2}} }>\mathrm{0} \\ $$$$\mathrm{f}''=\frac{\mathrm{2}}{\left(\mathrm{2}−\mathrm{x}\right)^{\mathrm{3}} }\geqslant\mathrm{0}\:\:\mathrm{f}\:\mathrm{convex} \\ $$$$\Rightarrow\mathrm{a}_{\mathrm{i}} \mathrm{f}\left(\mathrm{a}_{\mathrm{i}} \right)\geqslant\mathrm{f}\left(\Sigma\mathrm{a}_{\mathrm{i}} ^{\mathrm{2}} \right)…\mathrm{E} \\ $$$$\mathrm{Quadratic}\:\mathrm{Mean}\:\Rightarrow\sqrt[{\mathrm{2}}]{\frac{\Sigma\mathrm{a}_{\mathrm{i}} ^{\mathrm{2}} }{\mathrm{n}}}\geqslant\frac{\Sigma\mathrm{a}_{\mathrm{i}} }{\mathrm{n}} \\ $$$$\Sigma\mathrm{a}_{\mathrm{i}} ^{\mathrm{2}} \geqslant\frac{\mathrm{1}}{\mathrm{n}}…\mathrm{f}\:\mathrm{increase}\:\mathrm{function}\:\Rightarrow\mathrm{f}\left(\Sigma\mathrm{a}_{\mathrm{i}} ^{\mathrm{2}} \right)\geqslant\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{n}}\right) \\ $$$$\mathrm{E}\Rightarrow\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{a}_{\mathrm{i}} }{\mathrm{2}−\mathrm{a}_{\mathrm{i}} }\geqslant\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{n}}\right)=\frac{\mathrm{n}}{\mathrm{2n}−\mathrm{1}} \\ $$$$ \\ $$
Commented by Subhi last updated on 26/Jun/23
nice! (jensen′s inequality)
$${nice}!\:\left({jensen}'{s}\:{inequality}\right) \\ $$
Commented by witcher3 last updated on 27/Jun/23
thank You sir Yes
$$\mathrm{thank}\:\mathrm{You}\:\mathrm{sir}\:\mathrm{Yes} \\ $$

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