MathDB
Problems
Contests
National and Regional Contests
Switzerland Contests
Switzerland - Final Round
2012 Switzerland - Final Round
9
9
Part of
2012 Switzerland - Final Round
Problems
(1)
1+ab+bc+ca >= min { (a + b)^2 / ab } for a,b,c>0 with abc = 1
Source: Switzerland - 2012 Swiss MO Final Round p9
1/14/2023
Let
a
,
b
,
c
>
0
a, b, c > 0
a
,
b
,
c
>
0
be real numbers with
a
b
c
=
1
abc = 1
ab
c
=
1
. Show
1
+
a
b
+
b
c
+
c
a
≥
min
{
(
a
+
b
)
2
a
b
,
(
b
+
c
)
2
b
c
,
(
c
+
a
)
2
c
a
}
.
1 + ab + bc + ca \ge \min \left\{ \frac{(a + b)^2}{ab} , \frac{(b+c)^2}{bc} , \frac{(c + a)^2}{ca}\right\}.
1
+
ab
+
b
c
+
c
a
≥
min
{
ab
(
a
+
b
)
2
,
b
c
(
b
+
c
)
2
,
c
a
(
c
+
a
)
2
}
.
When does equality holds?
inequalities
algebra