A curious inequality concerning binomial coefficients
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Has anyone seen an inequality of this form before? It seems to be true (based on extensive testing), but I am not able to prove it.
Let $a_1,a_2,ldots,a_k$ be non-negative integers such that $sum_i a_i = A$. Then, for any non-negative integer $B le A$:
$$
sum_{(b_1,ldots,b_k): sum_i b_i = B} prod_i frac{binom{a_i}{b_i}}{binom{A-a_i}{B-b_i}} ge {binom{A}{B}}^{2-k}.
$$
The sum on the left is over all tuples $(b_1,b_2,ldots,b_k)$ of non-negative integers, with $b_i le a_i$ for all $i$, whose sum is equal to $B$.
inequalities binomial-coefficients
$endgroup$
add a comment |
$begingroup$
Has anyone seen an inequality of this form before? It seems to be true (based on extensive testing), but I am not able to prove it.
Let $a_1,a_2,ldots,a_k$ be non-negative integers such that $sum_i a_i = A$. Then, for any non-negative integer $B le A$:
$$
sum_{(b_1,ldots,b_k): sum_i b_i = B} prod_i frac{binom{a_i}{b_i}}{binom{A-a_i}{B-b_i}} ge {binom{A}{B}}^{2-k}.
$$
The sum on the left is over all tuples $(b_1,b_2,ldots,b_k)$ of non-negative integers, with $b_i le a_i$ for all $i$, whose sum is equal to $B$.
inequalities binomial-coefficients
$endgroup$
add a comment |
$begingroup$
Has anyone seen an inequality of this form before? It seems to be true (based on extensive testing), but I am not able to prove it.
Let $a_1,a_2,ldots,a_k$ be non-negative integers such that $sum_i a_i = A$. Then, for any non-negative integer $B le A$:
$$
sum_{(b_1,ldots,b_k): sum_i b_i = B} prod_i frac{binom{a_i}{b_i}}{binom{A-a_i}{B-b_i}} ge {binom{A}{B}}^{2-k}.
$$
The sum on the left is over all tuples $(b_1,b_2,ldots,b_k)$ of non-negative integers, with $b_i le a_i$ for all $i$, whose sum is equal to $B$.
inequalities binomial-coefficients
$endgroup$
Has anyone seen an inequality of this form before? It seems to be true (based on extensive testing), but I am not able to prove it.
Let $a_1,a_2,ldots,a_k$ be non-negative integers such that $sum_i a_i = A$. Then, for any non-negative integer $B le A$:
$$
sum_{(b_1,ldots,b_k): sum_i b_i = B} prod_i frac{binom{a_i}{b_i}}{binom{A-a_i}{B-b_i}} ge {binom{A}{B}}^{2-k}.
$$
The sum on the left is over all tuples $(b_1,b_2,ldots,b_k)$ of non-negative integers, with $b_i le a_i$ for all $i$, whose sum is equal to $B$.
inequalities binomial-coefficients
inequalities binomial-coefficients
asked 13 hours ago
Navin K.Navin K.
682
682
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1 Answer
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$begingroup$
By Cauchy–Bunyakovsky–Schwarz inequality we have
$$
left(sum prod_i frac{binom{a_i}{b_i}}{binom{A-a_i}{B-b_i}}right)left(sumprod_i binom{a_i}{b_i}binom{A-a_i}{B-b_i}right)geqslant
left(sum prod_ibinom{a_i}{b_i}right)^2=binom{A}{B}^2
.$$
Thus it suffices to prove that
$$
sumprod_i binom{a_i}{b_i}binom{A-a_i}{B-b_i}leqslant binom{A}B^k.
$$
But RHS is just the sum of the same guys without the restriction $sum b_i=B$.
$endgroup$
$begingroup$
Brilliant! Thank you.
$endgroup$
– Navin K.
5 hours ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
By Cauchy–Bunyakovsky–Schwarz inequality we have
$$
left(sum prod_i frac{binom{a_i}{b_i}}{binom{A-a_i}{B-b_i}}right)left(sumprod_i binom{a_i}{b_i}binom{A-a_i}{B-b_i}right)geqslant
left(sum prod_ibinom{a_i}{b_i}right)^2=binom{A}{B}^2
.$$
Thus it suffices to prove that
$$
sumprod_i binom{a_i}{b_i}binom{A-a_i}{B-b_i}leqslant binom{A}B^k.
$$
But RHS is just the sum of the same guys without the restriction $sum b_i=B$.
$endgroup$
$begingroup$
Brilliant! Thank you.
$endgroup$
– Navin K.
5 hours ago
add a comment |
$begingroup$
By Cauchy–Bunyakovsky–Schwarz inequality we have
$$
left(sum prod_i frac{binom{a_i}{b_i}}{binom{A-a_i}{B-b_i}}right)left(sumprod_i binom{a_i}{b_i}binom{A-a_i}{B-b_i}right)geqslant
left(sum prod_ibinom{a_i}{b_i}right)^2=binom{A}{B}^2
.$$
Thus it suffices to prove that
$$
sumprod_i binom{a_i}{b_i}binom{A-a_i}{B-b_i}leqslant binom{A}B^k.
$$
But RHS is just the sum of the same guys without the restriction $sum b_i=B$.
$endgroup$
$begingroup$
Brilliant! Thank you.
$endgroup$
– Navin K.
5 hours ago
add a comment |
$begingroup$
By Cauchy–Bunyakovsky–Schwarz inequality we have
$$
left(sum prod_i frac{binom{a_i}{b_i}}{binom{A-a_i}{B-b_i}}right)left(sumprod_i binom{a_i}{b_i}binom{A-a_i}{B-b_i}right)geqslant
left(sum prod_ibinom{a_i}{b_i}right)^2=binom{A}{B}^2
.$$
Thus it suffices to prove that
$$
sumprod_i binom{a_i}{b_i}binom{A-a_i}{B-b_i}leqslant binom{A}B^k.
$$
But RHS is just the sum of the same guys without the restriction $sum b_i=B$.
$endgroup$
By Cauchy–Bunyakovsky–Schwarz inequality we have
$$
left(sum prod_i frac{binom{a_i}{b_i}}{binom{A-a_i}{B-b_i}}right)left(sumprod_i binom{a_i}{b_i}binom{A-a_i}{B-b_i}right)geqslant
left(sum prod_ibinom{a_i}{b_i}right)^2=binom{A}{B}^2
.$$
Thus it suffices to prove that
$$
sumprod_i binom{a_i}{b_i}binom{A-a_i}{B-b_i}leqslant binom{A}B^k.
$$
But RHS is just the sum of the same guys without the restriction $sum b_i=B$.
answered 11 hours ago
Fedor PetrovFedor Petrov
50.8k6117233
50.8k6117233
$begingroup$
Brilliant! Thank you.
$endgroup$
– Navin K.
5 hours ago
add a comment |
$begingroup$
Brilliant! Thank you.
$endgroup$
– Navin K.
5 hours ago
$begingroup$
Brilliant! Thank you.
$endgroup$
– Navin K.
5 hours ago
$begingroup$
Brilliant! Thank you.
$endgroup$
– Navin K.
5 hours ago
add a comment |
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