Question

if a+b+c+d+e+f=1 then the maximum value of ab+bc+cd+de+ef . Given that a,b,c,d,e,f are positive...???

Answer 1

0

Answer 2

if none are negative then all must be 0 and a 1

Answer 3

for 0 .. hell no!!
and they r not necessary integers..

Answer 4

\[ 3(a^2+b^2+c^2) \leq (a+b+c)^2 \leq 3 (ab+bc+ac) \]
this might work

Answer 5

Woops!!
\[ 3(a^2+b^2+c^2) \geq (a+b+c)^2 \geq 3 (ab+bc+ac) \]

Answer 6

is it 1?

Answer 7

I think it must be less than 1

Answer 8

\[ (a+ b+c+d+e+f)^2 \geq 3(ab+bc+cd ..\\+de+ef+fa) \geq 3(ab+bc+cd+de+ef) \]
could be 1/3

Answer 9

a,b,c,d,e,f can any of them be 0?

Answer 10

not sure ... in that case 1/3 is the upper bound

Answer 11

1/4 ,,and am quite sure maybe ?

Answer 12

I also got 1/4 but I assumed except two all are zero.

Answer 13

exactly the same thing i did.. !!

Answer 14

WHAT DID U GET @shubhamsrg

Answer 15

well i didnt make much progress to be honest,,just 1/4 i got,,by putting 4 out of 6 =0 ,,and remaining then became = 1/2 each...

Answer 16

you can try the method of Lagrange multipliers if you like but this would be quite lengthy.

Answer 17

langrange gave me 1/8 as the solution,,not true,,ofcorse..

Answer 18

i won't try ...

Answer 19

the answer is 1/4 bt we need a mathematicaL proof..
we can take 2 no. to be approaching 1/2 and the rest all to be approaching 0 then the ans will approach zero...

Answer 20

unreachable

Answer 21

@Zekarias 1/4 is much much bigger than 5/36

Answer 22

(a+c+e)(b+d+f) less than (a+ b+...)/2 squared = 1/4 by AM/GM inequality.

Answer 23

It's easier once you have a target (ie 1/4)..

Answer 24

doesn't look like it has a maximum value

Answer 25

@estudier .... but (a+c+e)(b+d+f)=9 terms

Answer 26

As Zarkon says, you can only bound it.

So 6 terms still less than 9 terms, right..?

Answer 27

yes thats right ... so supremum of the expression being 1/4

Answer 28

Yes, I sort of did it backwards when I saw the 1/4...

Answer 29

what about the extra 4 terms wont they increase its value..