Question

Prove that if a/b=c/d a, b, c, d are positive numbers and a is the biggest prove that a+d>b+c.

Answer 1

@TheSmartOne @Loser66 @ash2326 @bibby @Ashleyisakitty

Answer 2

@TuringTest @Directrix

Answer 3

@freckles

Answer 4

so we have the following given:
\[\frac{a}{b}=\frac{c}{d}\]
\[a>b,c,d\]
\[a,b,c,d>0\]

Answer 5

yeah

Answer 6

well from the first equation we have \[ad=cb\]
... \[ad+dd=cb+dd\]
\[d(a+d)=cb+dd\]
\[d(a+d)+bb=cb+dd+bb\]
\[d(a+b)+bb=b(c+b)+dd\]
hmm... maybe we can think something along this idea
Like maybe we can alter this idea or something...
what do you think?

Answer 7

oops I changed d to b in the last line there*

Answer 8

wel let me think

Answer 9

and what then? seems a start is correct

Answer 10

you know ad=bc right?
and you know a>b
so aa>bb

Answer 11

yeah

Answer 12

so
what sign goes in between ad+aa ___ bc+bb

Answer 13

>

Answer 14

so we have ad+aa>bc+bb
or a(d+a)>b(c+b)

Answer 15

we are given a>b
do you think you can say anything else about the inequality a(d+a)>b(c+d) given that a>b

Answer 16

yeah

Answer 17

need to remove a and b from sides dont know how because they are firrent

Answer 18

by contradiction if we assume that this case
\(\large \begin{align} \color{black}{a<(b=c=d)\hspace{.33em}\\~\\
}\end{align}\)

then

\(\large \begin{align} \color{black}{a+d>b+c\hspace{.33em}\\~\\
a+d>d+c\hspace{.33em}\\~\\
a>c\hspace{.33em}\\~\\
}\end{align}\)
which contradicts our idea

Answer 19

but b c d are not equal

Answer 20

yes its only for that case

Answer 21

but for all cases?

Answer 22

@freckles we are nearly done but dont know last step

Answer 23

a*(d+a)=b*(c+b)
if we remove a and b we get what we need

Answer 24

woops >

Answer 25

lol my soln seems wrong since a/b=c/d given

Answer 26

right i was thinking
well if we have a/2=2/2
then a is also 2

Answer 27

so they are all equal is b=c=d

Answer 28

if b=c=d*

Answer 29

well 6/3=2/1

Answer 30

gono offline of a second

Answer 31

yeah back