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National and Regional Contests
Cuba Contests
Cuba MO
2001 Cuba MO
6
6
Part of
2001 Cuba MO
Problems
(1)
\frac{a}{a+c} + \frac{b}{b+a} > \frac{c}{b+c} , ax^2 - 4bx + 4c = 0
Source: - 2001 Cuba MO 2.6
9/15/2024
The roots of the equation
a
x
2
−
4
b
x
+
4
c
=
0
ax^2 - 4bx + 4c = 0
a
x
2
−
4
b
x
+
4
c
=
0
with
a
>
0
a > 0
a
>
0
belong to interval
[
2
,
3
]
[2, 3]
[
2
,
3
]
. Prove that: a)
a
≤
b
≤
c
<
a
+
b
.
a \le b \le c < a + b.
a
≤
b
≤
c
<
a
+
b
.
b)
a
a
+
c
+
b
b
+
a
>
c
b
+
c
.
\frac{a}{a+c} + \frac{b}{b+a} > \frac{c}{b+c} .
a
+
c
a
+
b
+
a
b
>
b
+
c
c
.
algebra
inequalities