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Problems
Contests
National and Regional Contests
Poland Contests
Poland - Second Round
1991 Poland - Second Round
1
prod a_i + prod b_i <= prod c_i + prod d_i
prod a_i + prod b_i <= prod c_i + prod d_i
Source: Polish MO Recond Round 1991 p1
September 9, 2024
algebra
inequalities
Product
Problem Statement
The numbers
a
i
a_i
a
i
,
b
i
b_i
b
i
,
c
i
c_i
c
i
,
d
i
d_i
d
i
satisfy the conditions
0
≤
c
i
≤
a
i
≤
b
i
≤
d
i
0\leq c_i \leq a_i \leq b_i \leq d_i
0
≤
c
i
≤
a
i
≤
b
i
≤
d
i
and
a
i
+
b
i
=
c
i
+
d
i
a_i+b_i = c_i+d_i
a
i
+
b
i
=
c
i
+
d
i
for
i
=
1
,
2
,
…
,
n
i=1,2 ,\ldots,n
i
=
1
,
2
,
…
,
n
. Prove that
∏
i
=
1
n
a
i
+
∏
i
=
1
n
b
i
≤
∏
i
=
1
n
c
i
+
∏
i
=
1
n
d
i
\prod_{i=1}^n a_i + \prod_{i=1}^n b_i \leq \prod_{i=1}^n c_i + \prod_{i=1}^n d_i
i
=
1
∏
n
a
i
+
i
=
1
∏
n
b
i
≤
i
=
1
∏
n
c
i
+
i
=
1
∏
n
d
i
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