MathDB
Problems
Contests
National and Regional Contests
Ukraine Contests
Official Ukraine Selection Cycle
Ukraine Team Selection Test
2013 Ukraine Team Selection Test
5
5
Part of
2013 Ukraine Team Selection Test
Problems
(1)
\sqrt[3]{(x+y)/2z}+\sqrt[3]{(y+z)/2y}+\sqrt[3]{(z+x)/2y} <= [5(x+y+z)+9]/8
Source: Ukraine TST 2013 p5
4/28/2020
For positive
x
,
y
x, y
x
,
y
, and
z
z
z
that satisfy the condition
x
y
z
=
1
xyz = 1
x
yz
=
1
, prove the inequality
x
+
y
2
z
3
+
y
+
z
2
x
3
+
z
+
x
2
y
3
≤
5
(
x
+
y
+
z
)
+
9
8
\sqrt[3]{\frac{x+y}{2z}}+\sqrt[3]{\frac{y+z}{2x}}+\sqrt[3]{\frac{z+x}{2y}}\le \frac{5(x+y+z)+9}{8}
3
2
z
x
+
y
+
3
2
x
y
+
z
+
3
2
y
z
+
x
≤
8
5
(
x
+
y
+
z
)
+
9
inequalities
algebra