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Inequality with x^2+y^2+z^2=2
Inequality with x^2+y^2+z^2=2
Source:
September 5, 2010
inequalities
algebra
Problem Statement
Prove that if
x
,
y
,
z
x, y, z
x
,
y
,
z
are real numbers such that
x
2
+
y
2
+
z
2
=
2
x^2+y^2+z^2 = 2
x
2
+
y
2
+
z
2
=
2
, then
x
+
y
+
z
≤
x
y
z
+
2.
x + y + z \leq xyz + 2.
x
+
y
+
z
≤
x
yz
+
2.
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