exercise 1.12[section-1,BOMC]

[photon]

let, $f(a,b,c,d)=(a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2$ where $a<b<c<d$,show that
$f(a,c,b,d)>f(a,b,c,d)>f(a,b,d,c)$

now ,$f(a,c,b,d)>f(a,b,c,d)$ will be true iff $ac+bc+bd+ad>ab+bc+cd+ad$
or,$ac-ab+bd-cd>0$ or,$(b-c)(d-a)>0$.but it is impossible as $(b-c)$ is negative,$(d-a)$ is positive.
so that should be $(b-c)(d-a)<0$.hence the problem shouldn't be proving $f(a,c,b,d)<f(a,b,c,d)<f(a,b,d,c)$????

[Hasib]

You make a littele mistake. It would be
$ac+bc+bd+ad<ab+bc+cd+ad$
cause, you give up all minus by multiple by $-1$ but forgot to change inequality sign.

[photon]

oops,silly mistake,thanks.

[nafistiham]

could it be proved like this?

first let us prove $f(a,c,b,d)>f(a,b,c,d)$
$a<d$
or,$-a>-d$
or,$a(b-c)>d(b-c)$
or,$ab-ac>bd-cd$
or,$ab+cd>ac+bd$
or,$2ab+2bc+2cd+2da>2ac+2bc+2bd+2da$
the later part is simple on hand writing.

plz state any mistake i have done. $f(a,b,c,d)>f(a,b,d,c)$ can be done in the same way.

[*Mahi*]

I don't think there is any. Noticing that the function is cyclic gives the second part directly.

[Labib]

I did it this way,

I substituted-
$d=a+k_1$;
$c=a+k_2$;
$b=a+k_3$
where, $k_1>k_2>k_3$ and simplified to deduce the solution.

[nafistiham]

here the combinations are thus, $a,b,c,d$ ;$a,b,c,d$ ;$a,d,b,c$
can there be any other combinations later?
the problem would be interesting if could be stated in a general form(i think the stating isnt easy enough )

$TIHAM$

[*Mahi*]

here the combinations are thus, $a,b,c,d$ ;$a,b,c,d$ ;$a,d,b,c$
can there be any other combinations later?
the problem would be interesting if could be stated in a general form(i think the stating isnt easy enough )

$TIHAM$

You can see from the function that it is cyclic, $f(d,c,b,a)=f(a,b,c,d)=f(b,c,d,a)$ and so on. So the $24$ combinations reduces to this $3$ cases.

[Labib]

Nice one, Mahi...

[*Mahi*]

Thanks for the complement...