<s>[INST] If $0 \leq a, b, c \leq 1$, the inequality $| a - b + c| \cdot |(a - c)^2 + b^2| \leq 1$ holds Please, help me to prove the inequality or find a counter-example
For $0 \leq a, b, c \leq 1$, prove that
$|a - b + c| \cdot |(a - c)^2 + b^2|  \leq 1$.
 [/INST] *

*$|a-c|\le 1 \implies(a-c)^2\le |a-c|$

*$b\le 1 \implies b^2\le b$
$\therefore (a-c)^2+b^2\le |a-c|+b=a-c+b=a+(b-c)$$\hspace{10mm}$(if $a\ge c$)
Now
\begin{align}
| a - b + c| \cdot |(a - c)^2 + b^2 | & = |[a - b + c] \cdot [(a - c)^2 + b^2] | \\[6pt]
& \le |[a - b + c] \cdot [a+(b-c)]| \\[6pt]
& = |[a - (b - c)] \cdot [a+(b-c)]| \\[6pt]
& = | a^2 - (b - c)^2| \\[6pt]
&\le 1 \text{ (because $a^2 \le 1;(b - c)^2\le1$ )}
\end{align}

Similarly, 
if $c\ge a$ we get $$ (a-c)^2+b^2\le |a-c|+b=c-a+b=c+(b-a)$$
Now
\begin{align}
| a - b + c| \cdot |(a - c)^2 + b^2 | & = |[a - b + c] \cdot [(a - c)^2 + b^2] | \\[6pt]
& \le |[a - b + c] \cdot [c+(b-a)]| \\[6pt]
& = |[c - (b - a)] \cdot [c+(b-a)]| \\[6pt]
& = | c^2 - (b - a)^2| \\[6pt]
&\le 1 \text{ (because $c^2 \le 1;(b - a)^2\le1$ )}
\end{align}


$\therefore$ In any case 
  $$|a - b + c| \cdot |(a - c)^2 + b^2| \le1$$

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