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Switzerland Contests
Switzerland - Final Round
2017 Switzerland - Final Round
10
10
Part of
2017 Switzerland - Final Round
Problems
(1)
1/(x + y + z)<= (x + y)(\sqrt3 z + 1) if x,y,z>=0 with xy + yz + zx = 1
Source: Switzerland - 2017 Swiss MO Final Round p10
1/14/2023
Let
x
,
y
,
z
x, y, z
x
,
y
,
z
be nonnegative real numbers with
x
y
+
y
z
+
z
x
=
1
xy + yz + zx = 1
x
y
+
yz
+
z
x
=
1
. Show that:
4
x
+
y
+
z
≤
(
x
+
y
)
(
3
z
+
1
)
.
\frac{4}{x + y + z} \le (x + y)(\sqrt3 z + 1).
x
+
y
+
z
4
≤
(
x
+
y
)
(
3
z
+
1
)
.
algebra
inequalities