A curious inequality concerning binomial coefficients
$begingroup$
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
asked 14 hours ago
Navin K.Navin K.
682
$endgroup$
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
asked 14 hours ago
Navin K.Navin K.
682
$endgroup$
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
asked 14 hours ago
Navin K.Navin K.
682
$endgroup$
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 14 hours ago
Navin K.Navin K.
682
asked 14 hours ago
Navin K.Navin K.
682
asked 14 hours ago
Navin K.Navin K.
682
asked 14 hours ago
Navin K.Navin K.
682
asked 14 hours ago
Navin K.Navin K.
682
682
add a comment |
add a comment |
1 Answer
1
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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 $sum b_i=B$.
answered 11 hours ago
Fedor PetrovFedor Petrov
50.9k6117233
$endgroup$
$begingroup$
Brilliant! Thank you.
$endgroup$
– Navin K.
6 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 $sum b_i=B$.
answered 11 hours ago
Fedor PetrovFedor Petrov
50.9k6117233
$endgroup$
$begingroup$
Brilliant! Thank you.
$endgroup$
– Navin K.
6 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 $sum b_i=B$.
answered 11 hours ago
Fedor PetrovFedor Petrov
50.9k6117233
$endgroup$
$begingroup$
Brilliant! Thank you.
$endgroup$
– Navin K.
6 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 $sum b_i=B$.
answered 11 hours ago
Fedor PetrovFedor Petrov
50.9k6117233
$endgroup$
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
50.9k6117233
answered 11 hours ago
Fedor PetrovFedor Petrov
50.9k6117233
answered 11 hours ago
Fedor PetrovFedor Petrov
50.9k6117233
answered 11 hours ago
Fedor PetrovFedor Petrov
50.9k6117233
50.9k6117233
$begingroup$
Brilliant! Thank you.
$endgroup$
– Navin K.
6 hours ago
add a comment |
$begingroup$
Brilliant! Thank you.
$endgroup$
– Navin K.
6 hours ago
$begingroup$
Brilliant! Thank you.
$endgroup$
– Navin K.
6 hours ago
$begingroup$
Brilliant! Thank you.
$endgroup$
– Navin K.
6 hours ago
add a comment |
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