If a > 0 and a<b<c<d,prove that if b+c = a+d then bc-ad>0.?

Question

Please show me complete and step by step working for 10marks.

Answer 1

Im sorry that instead of answering your homework I am giving you a dumb comment

Answer 2

Let's try a proof by contradiction. Our goal is to assume the conclusion to be false and to prove this CANNOT happen.
The negation of (bc - ac > 0) would be (bc - ad <= 0).

Assume bc - ad <= 0.
Let's add c^2 - d^2 to both sides, to obtain

bc + c^2 - ad - d^2 <= c^2 - d^2
c(b + c) - d(a + d) <= c^2 - d^2

Since it is given that b + c = a + d, we can replace the first set of brackets appropriately.

c(a + d) - d(a + d) <= c^2 - d^2

Low and behold, we can group the left hand side.

(c - d) (a + d) <= c^2 - d^2

Plus, we can factor the right hand side as a difference of squares.

(c - d) (a + d) <= (c - d) (c + d)

Remember that (c - d) is a negative number, since it is given that c < d, which would follow that c - d < 0. Whenever we divide an inequality by a negative number, the sign changes. That's what we're going to do; divide by (c - d) both sides.

a + d >= c + d

Subtract d both sides, to get

a >= c
However, it is given that a < c (since a < b < c < d). That means both
a >= c and a < c.

THIS IS A CONTRADICTION. (a can't both be greater than or equal to c, AND less than c).

Therefore, for a > 0 and a < b < c < d, if b+c = a+d, then
bc - ad > 0.

Answer 3

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Answer 4

heres what i did, a=1 b=2 c=3 d=4, so 1<2<3<4, 2+3 (bc) 1+4 (ad) so 5 (bc) - 5 (ad) equals 0 so its not greater, its equal. but when a=5 b=7 c=12 d=19 then bc-ac is greater than 0

Answer 5

From b+c = a+d you can find that

d-c = b-a

Now, all you have to do is rewrite bc - ad in terms of only two variables b,c.

bc -ad = bc-[b-(b-a)][c+(d-c)] since b-(b-a) = a & c+(d-c) =d
= bc-[b-(b-a)][c+(b-a)] from d-c = b-a above
= bc-[bc +b(b-a)-c(b-a)-(b-a)^2]
= bc-[bc+(b-a)(b-c) -(b-a)^2]
= -(b-a)(b-c) + (b-a)^2
= (b-a)[(b-a)-(b-c)]
= (b-a)(c-a)>0
Well, the last expression is positive since both factors
(b-a) and (c-a) are positive because a
Thus, bc-ad >0

Answer 6

a < b < c < d
(d-a) > (c -b) >0


square both sides

a^2 +d^2 - 2ad > b^2 + c^2 - 2bc .... 1
from the second (b+c) = (a+d) square both sides

b^2 + c^2 + 2bc = a^2 + d^2 + 2ad .... 2
add 1 and 2

a^2+b^2+c^2+d^2+ 2(bc-ad) > b^2+c^2+a^2+d^2 + 2(ad-bc)

2(bc-ad) > 2(ad-bc)
or 4(bc-ad) > 0
or bc-ad > 0
QED

Answer 7

Write
bc-ad = b(c+b-b) -a(a+d-a) = b(c+b) -b^2 -a(a+d) +a^2
= b(c+b) - b^2 -a(b+c) +a^2
substitute a+d=b+c
= (b-a)(c+b) - b^2 +a^2
=(b-a)*(c+b) -(b^2-a^2) = (b-a)*(c+b) - (b-a)(b+a)
= (b-a)* (c+b-(b+a)) = (b-a)(c-a) = product of two positive terms

Answer 8

This one's easy, but your math teacher wouldn't like it that you're cheating on your homework. Just try assigning some numbers to each of the symbols.