Given that $\displaystyle a+b\le c+1$, $\displaystyle b+c \le a+1$, $\displaystyle c+a\le b+1$ for the numbers $\displaystyle a,b,c \ge0$, show that $\displaystyle a^2+b^2+c^2\le2abc+1$.
Notice that the inequality we need to prove contains a term in $\displaystyle abc$. Thus we have to try somehow or other to bring that term into our working.
Now, since we are working with non-negative numbers, we can multiply all the inequalities together. Let’s try that.
$\displaystyle (a+b)(b+c)(c+a)\ \leq\ (a+1)(b+1)(c+1)$
The LHS is equal to $\displaystyle (ab+bc+ca)(a+b+c)-abc$ (expand them out and check). The RHS is equal to $\displaystyle abc+(ab+bc+ca)+(a+b+c)+1$. Hence
$\displaystyle (ab+bc+ca)(a+b+c)\ \leq\ 2abc+(ab+bc+ca)+(a+b+c)+1$
Which doesn’t bring us any closer to a solution, but at least we now have the $\displaystyle abc$ term to work with.
I'm sure this is supposed to be a slick application of one of those Jensen-type inequalities that they drill math olympiad candidates with. Here's a much more pedestrian approach (aesthetically unappealing because it breaks the symmetry).
If $\displaystyle a+b\leqslant c+1$ and $\displaystyle c+a\leqslant b+1$ then $\displaystyle b-c\leqslant 1-a$ and $\displaystyle c-b\leqslant 1-a$. Thus $\displaystyle |b-c|\leqslant 1-a$. In particular, $\displaystyle 1-a\geqslant0$. Also, $\displaystyle (b-c)^2\leqslant (1-a)^2$. Then
. . . . . .$\displaystyle \begin{array}{rcl}a^2+b^2+c^2 -2abc &=&
a^2 + \frac12(1+a)(b-c)^2 + \frac12(1-a)(b+c)^2 \vspace{1ex} \\
&\leqslant&a^2 + \frac12((1+a)(1-a)^2 + \frac12(1-a)(1+a)^2 \vspace{1ex} \\ &=&a^2 + \frac12((1-a^2)(1-a) + \frac12(1-a^2)(1+a) \vspace{1ex} \\
&=& a^2 + 1 - a^2 = 1.\end{array}$