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P3
ASU 543 All Soviet Union MO 1991 (x+y+z)^2/3>=x\sqrt{yz}+y\sqrt{zx}+z\sqrt{xy}
ASU 543 All Soviet Union MO 1991 (x+y+z)^2/3>=x\sqrt{yz}+y\sqrt{zx}+z\sqrt{xy}
Source:
August 14, 2019
algebra
inequalities
Problem Statement
Show that
(
x
+
y
+
z
)
2
3
≥
x
y
z
+
y
z
x
+
z
x
y
\frac{(x + y + z)^2}{3} \ge x\sqrt{yz} + y\sqrt{zx} + z\sqrt{xy}
3
(
x
+
y
+
z
)
2
≥
x
yz
+
y
z
x
+
z
x
y
for all non-negative reals
x
,
y
,
z
x, y, z
x
,
y
,
z
.
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