Well, Given these
conditions:
A + B < C + D
B + D
< A + C
A + D < B + C
By the
property of inequalities, same signed inqualities can add or subtract each
other
by adding the first
two
A+ 2B +D < A+ 2C +
D
subtract A+D from both sides, we
have
2B<2C
B<C
add
the last two equations together, we
have:
A+B+2D<A+B+2C
subtract
A+B from both sides, we
have:
2D<2C
D<C
ok,
without evaluating A and C's relationship, we know that B and D are out of the
game
add the first and last
together
2A+B+D<2C+B+D
Subtract
b+d from both
sides
2A<2C
A<C
the
winner is C
C is the biggest number among the
four given the conditions
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