APMO 1990 (Inequality with Combi)

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sowmitra
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APMO 1990 (Inequality with Combi)

Unread post by sowmitra » Wed Aug 08, 2012 8:40 pm

Let, $a_1,a_2,...,a_n$ be positive real numbers, and let $S_k$ be the sum of the products of $a_1,a_2,...,a_n$ taken $k$ at a time. Show that,
\[\displaystyle S_kS_{n-k}\geq \binom{n}{k}^2a_1a_2...a_n\]
\[\forall k=1,2,...,n-1\]
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sm.joty
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Re: APMO 1990 (Inequality with Combi)

Unread post by sm.joty » Thu Aug 09, 2012 12:57 pm

Well here is my solution
Let prove a quick lemma
Lemma:For finite n element set, the product of $a_{1},a_{2}.......,a_{n}$ taken $k$ at a time is
$[a_{1}a_{2}.......a_{n}]^{r}$
where $r_{1}=\displaystyle \binom{n-1}{k-1}$
because here every positive integer repeated for $r_{1}$ times.

Now back to main problem,
let, $x=\displaystyle \binom{n}{k}$
Using AM-GM inequality ,
$\frac{S_{n}}{x}\geq [a_{1}a_{2}.......a_{n}]^{\frac{r_{1}}{x}}$

if $r_{2}=\displaystyle \binom{n-1}{n-k-1}$
and $x=\displaystyle \binom{n}{k}=\displaystyle \binom{n}{n-k}$

then
$\frac{S_{n-k}}{x} \geq [a_{1}a_{2}.......a_{n}]^{\frac{r_{2}}{x}} $
Now
$\frac{r_{1}}{x}=\frac{k}{n}$ and
$\frac{r_{2}}{x}=\frac{n-k}{n}$
so $\frac{r_{1}+r_{2}}{x}=1$

So, multiplying both inequality we get,
$S_{n}S_{n-k}\geq \displaystyle \binom{n}{k}^{2}[a_{1}a_{2}.......a_{n}]$

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sowmitra
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Re: APMO 1990 (Inequality with Combi)

Unread post by sowmitra » Thu Aug 09, 2012 5:49 pm

Vaia, good use of the AM-GM inequality. Thanks for the post. :)
"Rhythm is mathematics of the sub-conscious."
Some-Angle Related Problems;

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