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Baltic Way
2023 Baltic Way
5
Two variable inequality
Two variable inequality
Source: Baltic Way 2023/5
November 11, 2023
inequalities
Problem Statement
Find the smallest positive real
α
\alpha
α
, such that
x
+
y
2
≥
α
x
y
+
(
1
−
α
)
x
2
+
y
2
2
\frac{x+y} {2}\geq \alpha\sqrt{xy}+(1 - \alpha)\sqrt{\frac{x^2+y^2}{2}}
2
x
+
y
≥
α
x
y
+
(
1
−
α
)
2
x
2
+
y
2
for all positive reals
x
,
y
x, y
x
,
y
.
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