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Problem
Problem
If
ln
a
b
−
c
=
ln
b
c
−
a
=
ln
c
a
−
b
\frac{\ln a}{b-c} = \frac{\ln b}{c-a} = \frac{\ln c}{a-b}
b
−
c
l
n
a
=
c
−
a
l
n
b
=
a
−
b
l
n
c
, prove the following. (i)
a
b
c
=
1
abc = 1
ab
c
=
1
(ii)
a
a
⋅
b
b
⋅
c
c
=
1
a^a \cdot b^b \cdot c^c = 1
a
a
⋅
b
b
⋅
c
c
=
1
(iii)
a
b
2
+
b
c
+
c
2
⋅
b
c
2
+
c
a
+
a
2
⋅
c
a
2
+
a
b
+
b
2
=
1
a^{b^2+bc+c^2} \cdot b^{c^2+ca+a^2} \cdot c^{a^2+ab+b^2} = 1
a
b
2
+
b
c
+
c
2
⋅
b
c
2
+
c
a
+
a
2
⋅
c
a
2
+
ab
+
b
2
=
1
(iv)
a
+
b
+
c
≥
3
a + b + c \ge 3
a
+
b
+
c
≥
3
(v)
a
a
+
b
b
+
c
c
≥
3
a^a + b^b + c^c \ge 3
a
a
+
b
b
+
c
c
≥
3
(vi)
a
b
2
+
b
c
+
c
2
+
b
c
2
+
c
a
+
a
2
+
c
a
2
+
a
b
+
b
2
≥
3
a^{b^2+bc+c^2} + b^{c^2+ca+a^2} + c^{a^2+ab+b^2} \ge 3
a
b
2
+
b
c
+
c
2
+
b
c
2
+
c
a
+
a
2
+
c
a
2
+
ab
+
b
2
≥
3
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