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Prove inequality

mathworker

Active member
May 31, 2013
118
IMO-2012:
let \(\displaystyle a_2,a_3,.....,a_n\) be positive real numbers that satisfy a2.a3......a​n=1 .Prove that,
\(\displaystyle (a_2+1)^2.(a_3+1)^3.......(a_n+1)^n>n^n\)
hint:
Use A.M>G.M
 

mathworker

Active member
May 31, 2013
118
Re: prove inequality

Look at this only after giving a serious try
hint#2
split the terms in (ak+1) into k terms and apply AM>GM
 

mathworker

Active member
May 31, 2013
118
OFICIAL answer by IMO:
\(\displaystyle (a_k+1)=(a_k+\frac{1}{k-1}+\frac{1}{k-1}........\frac{1}{k-1})\)(k-1) times
Apply AM>GM
\(\displaystyle (a_k+1)^k>k^k\frac{a_k}{(k-1)^{k-1}}\)
therefore,
\(\displaystyle \prod_{k=2}^{n} (a_k+1)^k>\prod_{k=2}^{n}k^k\frac{a_k}{(k-1)^{k-1}}\)
\(\displaystyle \prod_{k=2}^{n} (a_k+1)^k>2^2*\frac{a_2}{1^{1}}*3^3*\frac{a_3}{2^{2}}......n^n*\frac{a_{n}}{(n-1)^{n-1}}\)
\(\displaystyle \prod_{k=2}^{n} (a_k+1)^k>\frac{\cancel{2^2}}{1^{1}}\frac{\cancel{3^3}}{\cancel{2^{2}}}......\frac{n^n}{\cancel{(n-1)^{n-1}}}({a_2}.{a_3}.......{a_{n}})\)
\(\displaystyle \prod_{k=2}^{n} (a_k+1)^k>n^n.(1)\)