Mathematical Reflections 2011, Issue 6 - Problem J214

Problem:
Let $a,b,c,d,e \in [1,2]$. Prove that
$$ab+bc+cd+de+ea \geq a^2+b^2+c^2+d^2-e^2.$$

Proposed by Ion Dobrota and Adrian Zahariuc.

Solution:

It is equivalent to prove that given $a,b,c,d,e \in [1,2]$, the following inequality holds
$$(a-b)^2+(b-c)^2+(c-d)^2+(d-e)^2+(e-a)^2 \leq 4e^2.$$
Clearly,
$$(a-b)^2 \leq 1, \qquad (b-c)^2 \leq 1, \qquad (c-d)^2 \leq 1, \qquad (d-e)^2 \leq 1, \qquad (e-a)^2 \leq 1,$$
so
$$(a-b)^2+(b-c)^2+(c-d)^2+(d-e)^2+(e-a)^2 \leq |a-b|+|b-c|+|c-d|+|d-e|+|e-a|.$$
The right hand side is maximum if and only if there is only one term which cancels out when removing the absolute values. Without loss of generality, we can suppose that $b$ cancels out, so
$$|a-b|+|b-c|+|c-d|+|d-e|+|e-a| \leq 2(a+d)-2(c+e) \leq 4 \leq 4e^2,$$
and the desired conclusion follows. From the made observations, we have four cases for the equality:

(i) If $a$ cancels out, then $(b-a)+(b-c)+(d-c)+(d-1)+(a-1)=2(b+d)-2c-2=4 \iff 4 \leq b+d \iff b=d=2, c=1$, and $a=1,2$.

(ii) If $b$ cancels out, then $(a-b)+(b-c)+(d-c)+(d-1)+(a-1)=2(a+d)-2c-2=4 \iff 4 \leq a+d \iff a=d=2, c=1$, and $b=1,2$.

(iii) If $c$ cancels out, then $(a-b)+(c-b)+(d-c)+(d-1)+(a-1)=2(a+d)-2b-2=4 \iff 4 \leq a+d \iff a=d=2, b=1$, and $c=1,2$.

(iv) If $d$ cancels out, then $(a-b)+(c-b)+(c-d)+(d-1)+(a-1)=2(a+c)-2b-2=4 \iff 4 \leq a+c \iff a=c=2, b=1$, and $d=1,2$.

In conclusion, the equality can be obtained if $e=1$ and occurs if and only if $$(a,b,c,d,e) \in \{(1,2,1,2,1),(2,2,1,2,1),(2,1,1,2,1),(2,1,2,2,1),(2,1,2,1,1)\}.$$