, 2007

Adding the first two, we get $2b \le 2$ so that $b \le 1$. Similarly, we get $c \le 1$ and $a \le 1$. Put $\alpha = 1 - a$, $\beta = 1 - b$ and $\gamma = 1 - c$. Then $0 \le \alpha, \beta, \gamma \le 1$ and
$$\alpha + \beta\gamma, \quad \beta + \gamma \ge \alpha, \quad \gamma + \alpha \ge \alpha.$$
The inequality to be proved reduces to
$$\alpha^2 + \beta^2 + \gamma^2 \le 2(\alpha\beta + \beta\gamma + \gamma\alpha) - 2\alpha\beta\gamma.$$
We may assume that $\gamma$ is the largest among the three, so that $\alpha \le \gamma$ and $\beta \le \gamma$. Using $\gamma \le \alpha + \beta$, we get $\gamma^2 \le \gamma(\alpha + \beta)$. Also $\alpha^2 \le \alpha\gamma$ and $\beta^2 \le \beta\gamma$. Thus $\alpha^2 + \beta^2 + \gamma^2 \le 2(\alpha\gamma + \beta\gamma)$. Hence it suffices to prove that $2\alpha\beta\gamma \le 2\alpha\beta$. This follows from $\gamma \le 1$.

(Amar Arpit Goel, Utkarsh Tripati). As in the first solution, we conclude that $0 \le a, b, c \le 1$. The symmetry shows that we may assume $a \ge b \ge c$. We may write the inequality in the form
$$(a-b)^2 \le (1-c)(1+c-2ab).$$
Observe that $1-c \ge a-b \ge 0$, by $b+c \le a+1$. Using $1+c \ge a+b$, we get $1+c \ge a+b \ge a+ab \ge a(1+b)+b(a-1)$, since $b \ge 0$ and $a \le 1$. Thus $1+c-2b \ge a-b \ge 0$. It follows that
$$(a-b)^2 \le (1-c)(1+c-2ab),$$
as required.