-1
$\begingroup$

Let $a,b,c, d$ be real numbers such that $a+b+c+d=0$ and $a^2+b^2+c^2+d^2=1$. Then the smallest possible value of expression $(a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2$ lies in the interval:

i)$(0-1.5)$

ii)$(1.5-2.5)$

iii)$(2.5-3)$

iv)$(3.5-4)$

$\endgroup$
1
  • $\begingroup$ This is hard to read. Try to read the tutorial on formatting for this site. $\endgroup$ Commented Oct 18, 2018 at 23:31

1 Answer 1

1
$\begingroup$

\begin{align} I &= (a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2 \\ &= 2(a^2+b^2+c^2+d^2)-2(ab+bc+cd+da) \\ &= 2-2(a+c)(b+d) \\ &= 2+2(a+c)^2 \ge 2 \end{align}

we can get equality at $a=b=1/2,c=d=-1/2$

$\endgroup$
4
  • $\begingroup$ we can also find the maximum: $I=2+(a+c)^2+(b+d)^2=3+2(ac+bd)\le3+a^2+c^2+b^2+d^2=4$ $\endgroup$ Commented Oct 19, 2018 at 0:03
  • $\begingroup$ How 2-2(a+c)^2>=2 came? Please explain. $\endgroup$
    – user585143
    Commented Oct 19, 2018 at 1:17
  • $\begingroup$ @user585143 it's $2+2(a+c)^2 \ge 2$, the plus sign comes from $b+d=-(a+c)$. $\endgroup$ Commented Oct 19, 2018 at 1:19
  • $\begingroup$ Thank you now it is easily understandable $\endgroup$
    – user585143
    Commented Oct 20, 2018 at 3:01

Not the answer you're looking for? Browse other questions tagged .