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National and Regional Contests
Serbia Contests
Federal Math Competition of Serbia and Montenegro
2005 Federal Math Competition of S&M
Problem 1
sum x/(√(y+z))>=√(3/2(x+y+z)) in R+
sum x/(√(y+z))>=√(3/2(x+y+z)) in R+
Source: Serbia MO 2005 3&4th Grades P1
April 11, 2021
Inequality
inequalities
Problem Statement
If
x
,
y
,
z
x,y,z
x
,
y
,
z
are positive numbers, prove that
x
y
+
z
+
y
z
+
x
+
z
x
+
y
≥
3
2
(
x
+
y
+
z
)
.
\frac x{\sqrt{y+z}}+\frac y{\sqrt{z+x}}+\frac z{\sqrt{x+y}}\ge\sqrt{\frac32(x+y+z)}.
y
+
z
x
+
z
+
x
y
+
x
+
y
z
≥
2
3
(
x
+
y
+
z
)
.
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