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National and Regional Contests
Romania Contests
District Olympiad
2023 District Olympiad
P3
Inequality
Inequality
Source: Romanian District Olympiad 2023 9.3
March 11, 2023
algebra
inequalities
Problem Statement
Let
x
,
y
x,y{}
x
,
y
and
z
z{}
z
be positive real numbers satisfying
x
+
y
+
z
=
1
x+y+z=1
x
+
y
+
z
=
1
. Prove that [*]
1
−
x
2
−
y
z
x
2
+
x
=
(
1
−
y
)
(
1
−
z
)
x
2
+
x
;
1-\frac{x^2-yz}{x^2+x}=\frac{(1-y)(1-z)}{x^2+x};
1
−
x
2
+
x
x
2
−
yz
=
x
2
+
x
(
1
−
y
)
(
1
−
z
)
;
[*]
x
2
−
y
z
x
2
+
x
+
y
2
−
z
x
y
2
+
y
+
z
2
−
x
y
z
2
+
z
⩽
0.
\frac{x^2-yz}{x^2+x}+\frac{y^2-zx}{y^2+y}+\frac{z^2-xy}{z^2+z}\leqslant 0.
x
2
+
x
x
2
−
yz
+
y
2
+
y
y
2
−
z
x
+
z
2
+
z
z
2
−
x
y
⩽
0.
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