# Which of the numbers A, B, C, D is the largest number, if the following threeConditions are meet. A + B < C + DB + D < A + C A + D < B + C

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A + B < C + D

B + D < A + C

A + D < B + C

A < C + D - B

B + D - C < A

Therefore...

B + D - C < C + D - B

B - C < C - B

2B < 2C

B < C

A + B < C + D

B + D < A + C

A + D < B + C

B < C + D - A

A + D - C < B

Therefore...

A + D - C < C + D - A

A - C < C - A

2A < 2C

A < C

A + B < C + D

B + D < A + C

A + D < B + C

A and B are both less than C. C is also greater than D when either is added to A or B.

**Therefore, C is the largest number.**

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**