, 2003

First Solution. (Based on work by Matthew Tang and Anders Kaseorg)
By multiplying $a$, $b$, and $c$ by a suitable factor, we reduce the problem to the case when $a + b + c = 3$. The desired inequality reads
$$\frac{(a+3)^2}{2a^2+(3-a)^2} + \frac{(b+3)^2}{2b^2+(3-b)^2} + \frac{(c+3)^2}{2c^2+(3-c)^2} \le 8.$$
Set
$$f(x) = \frac{(x+3)^2}{2x^2 + (3-x)^2}$$
It suffices to prove that $f(a) + f(b) + f(c) \le 8$. Note that
$$\begin{aligned} f(x) &= \frac{x^2 + 6x + 9}{3(x^2 - 2x + 3)} = \frac{1}{3} \cdot \frac{x^2 + 6x + 9}{x^2 - 2x + 3} \\ &= \frac{1}{3} \left( 1 + \frac{8x + 6}{x^2 - 2x + 3} \right) \\ &= \frac{1}{3} \left( 1 + \frac{8x + 6}{(x-1)^2 + 2} \right) \le \frac{1}{3} \left( 1 + \frac{8x + 6}{2} \right) \\ &= \frac{1}{3}(4x + 4). \end{aligned}$$
Hence,
$$f(a) + f(b) + f(c) \le \frac{1}{3}(4a + 4 + 4b + 4 + 4c + 4) = 8,$$
as desired, with equality if and only if $a = b = c$.

Second Solution. (By Liang Qin) Setting $x = a+b$, $y = b+c$, $z = c+a$ gives $2a+b+c = x+z$, hence $2a = x+z-y$ and their analogous forms. The desired inequality becomes
$$\frac{2(x+z)^2}{(x+z-y)^2+2y^2} + \frac{2(z+y)^2}{(z+y-x)^2+2x^2} + \frac{2(y+x)^2}{(y+x-z)^2+2z^2} \le 8.$$
Because $2(s^2 + t^2) \ge (s + t)^2$ for all real numbers $s$ and $t$, we have $2(x + z - y)^2 + 2y^2 \ge (x + z - y + y)^2 = (x + z)^2$. Hence
$$\begin{aligned} \frac{2(x+z)^2}{(x+z-y)^2+2y^2} &= \frac{4(x+z)^2}{2(x+z-y)^2+4y^2} \le \frac{4(x+z)^2}{(x+z)^2+2y^2} \\ &= \frac{4}{1+2 \cdot \frac{y^2}{(x+z)^2}} \le \frac{4}{1+2 \cdot \frac{y^2}{2(x^2+z^2)}} \\ &= \frac{4(x^2+z^2)}{x^2+y^2+z^2}. \end{aligned}$$
It is not difficult to see that the desired result follows from summing up the above inequality and its analogous forms.

Third Solution. (By Richard Stong) Note that
$$\begin{aligned} (2x + y)^2 + 2(x - y)^2 &= 4x^2 + 4xy + y^2 + 2x^2 - 4xy + 2y^2 \\ &= 3(2x^2 + y^2). \end{aligned}$$
Setting $x = a$ and $y = b + c$ yields
$$(2a + b + c)^2 + 2(a - b - c)^2 = 3(2a^2 + (b+c)^2).$$
Thus, we have
$$\begin{aligned} \frac{(2a + b + c)^2}{2a^2 + (b + c)^2} &= \frac{3(2a^2 + (b+c)^2) - 2(a-b-c)^2}{2a^2 + (b+c)^2} \\ &= 3 - \frac{2(a-b-c)^2}{2a^2 + (b+c)^2}. \end{aligned}$$
and its analogous forms. Thus, the desired inequality is equivalent to
$$\frac{(a - b - c)^2}{2a^2 + (b + c)^2} + \frac{(b - a - c)^2}{2b^2 + (c + a)^2} + \frac{(c - a - b)^2}{2c^2 + (a + b)^2} \geq \frac{1}{2}.$$
Because $(b+c)^2 \le 2(b^2+c^2)$, we have $2a^2+(b+c)^2 \le 2(a^2+b^2+c^2)$ and its analogous forms. It suffices to show that
$$\frac{(a - b - c)^2}{2(a^2 + b^2 + c^2)} + \frac{(b - a - c)^2}{2(a^2 + b^2 + c^2)} + \frac{(c - a - b)^2}{2(a^2 + b^2 + c^2)} \ge \frac{1}{2},$$
or,
$$(a - b - c)^2 + (b - a - c)^2 + (c - a - b)^2 \ge a^2 + b^2 + c^2.$$
Multiplying this out, the left-hand side of the last inequality becomes $3(a^2+b^2+c^2)-2(ab+bc+ca)$. Therefore the last inequality is equivalent to $2[a^2 + b^2 + c^2 - (ab + bc + ca)] \ge 0$, which is evident because
$$2[a^2 + b^2 + c^2 - (ab + bc + ca)] = (a - b)^2 + (b - c)^2 + (c - a)^2.$$
Equalities hold if and only if $(b+c)^2 = 2(b^2+c^2)$ and $(c+a)^2 = 2(c^2+a^2)$, that is, $a = b = c$.

Fourth Solution. We first convert the inequality into
$$\frac{2a(a + 2b + 2c)}{2a^2 + (b + c)^2} + \frac{2b(b + 2c + 2a)}{2b^2 + (c + a)^2} + \frac{2c(c + 2a + 2b)}{2c^2 + (a + b)^2} \le 5.$$
Splitting the 5 among the three terms yields the equivalent form
$$\sum_{\text{cyc}} \frac{4a^2 - 12a(b + c) + 5(b + c)^2}{3[2a^2 + (b + c)^2]} \ge 0, \quad (1)$$
where $\sum_{\text{cyc}}$ is the cyclic sum of variables $(a, b, c)$. The numerator of the term shown factors as $(2a-x)(2a-5x)$, where $x = b+c$. We will show
$$\frac{(2a-x)(2a-5x)}{3(2a^2+x^2)} \geq -\frac{4(2a-x)}{3(a+x)}. \qquad (2)$$
Indeed, (2) is equivalent to
$$(2a-x)[(2a-5x)(a+x) + 4(2a^2+x^2)] \geq 0,$$
which reduces to
$$(2a-x)(10a^2 - 3ax - x^2) = (2a-x)^2(5a+x) \geq 0,$$
which is evident. We proved that
$$\frac{4a^2 - 12a(b+c) + 5(b+c)^2}{3[2a^2 + (b+c)^2]} \geq -\frac{4(2a-b-c)}{3(a+b+c)},$$
hence (1) follows. Equality holds if and only if $2a = b + c$, $2b = c + a$, $2c = a + b$, i.e., when $a = b = c$.

Fifth Solution. Given a function $f$ of $n$ variables, we define the symmetric sum
$$\sum_{\text{sym}} f(x_1, \dots, x_n) = \sum_{\sigma} f(x_{\sigma(1)}, \dots, x_{\sigma(n)})$$
where $\sigma$ runs over all permutations of $1, \dots, n$ (for a total of $n!$ terms). For example, if $n = 3$, and we write $x, y, z$ for $x_1, x_2, x_3$,
$$\begin{aligned} \sum_{\text{sym}} x^3 &= 2x^3 + 2y^3 + 2z^3 \\ \sum_{\text{sym}} x^2y &= x^2y + y^2z + z^2x + x^2z + y^2x + z^2y \\ \sum_{\text{sym}} xyz &= 6xyz. \end{aligned}$$
We combine the terms in the desired inequality over a common denominator and use symmetric sum notation to simplify the algebra. The numerator of the difference between the two sides is
$$2 \sum_{\text{sym}} (4a^6 + 4a^5b + a^4b^2 + 5a^4bc + 5a^3b^3 - 26a^3b^2c + 7a^2b^2c^2), \quad (3)$$
and it suffices to show the the expression in (3) is always greater or equal to 0. By the Weighted AM-GM Inequality, we have $4a^6 + b^6 + c^6 \geq 6a^4bc$, $3a^5b + 3a^5c + b^5a + c^5a \geq 8a^4bc$, and their analogous forms. Adding those inequalities yields
$$\sum_{\text{sym}} 6a^6 \geq \sum_{\text{sym}} 6a^4bc \quad \text{and} \quad \sum_{\text{sym}} 8a^5b \geq \sum_{\text{sym}} 8a^4bc.$$
Consequently, we obtain
$$\sum_{\text{sym}} 4a^6 + 4a^5b + 5a^4bc \geq \sum_{\text{sym}} 13a^4bc. \quad (4)$$
Again by the AM-GM Inequality, we have $a^4b^2 + b^4c^2 + c^4a^2 \geq 4a^2b^2c^2$, $a^3b^3 + b^3c^3 + c^3a^3 \geq 3a^2b^2c^2$, and their analogous forms. Thus,
$$\sum_{\text{sym}} a^4b^2 + 5a^3b^3 \geq \sum_{\text{sym}} 6a^2b^2c^2,$$
or
$$\sum_{\text{sym}} a^4b^2 + 5a^3b^3 + 7a^2b^2c^2 \geq \sum_{\text{sym}} 13a^2b^2c^2. \quad (5)$$
Recalling Schur's Inequality, we have
$$\begin{aligned} & a^3 + b^3 + c^3 + 3abc - (a^2b + b^2c + c^2a + ab^2 + bc^2 + ca^2) \\ & = a(a-b)(a-c) + b(b-a)(b-c) + c(c-a)(c-b) \geq 0, \end{aligned}$$
or
$$\sum_{\text{sym}} a^3 - 2a^2b + abc \geq 0.$$
Thus
$$\sum_{\text{sym}} 13a^4bc - 26a^3b^2c + 13a^2b^2c^2 \geq 13abc \sum_{\text{sym}} a^3 - 2a^2b + abc \geq 0. \quad (6)$$
Adding (4), (5), and (6) yields (3).