Irish Mathematical Olympiad

Consider the inequality on the left. We prove this using the triangle inequality according to which $a < b + c$, $b < c + a$, and $c < a + b$. Applying this,
$$\begin{aligned} & (a + b + c)(a^2 + b^2 + c^2) - 2(a^3 + b^3 + c^3) \\ &= a^2(b + c) + b^2(c + a) + c^2(a + b) - a^3 - b^3 - c^3 \\ &= a^2(b + c - a) + b^2(c + a - b) + c^2(a + b - c) > 0. \end{aligned}$$

The inequality on the right holds more generally for any triple of positive numbers $a$, $b$, $c$, and can be handled by Hölder's inequality, or by strict convexity of the function $t \mapsto t^p$ with $p \ge 1$. Either of these tells us that: if $x$, $y$, $z > 0$ and $p \ge 1$, then
$$(x + y + z) \le (x^p + y^p + z^p)^{1/p} 3^{(p-1)/p} \quad \text{or equivalently} \\ \left(\frac{x+y+z}{3}\right)^p \le \frac{x^p+y^p+z^p}{3},$$
with equality iff $x = y = z$. Apply this first with $x = a$, $y = b$, $z = c$ and $p = 3$, to get
$$a+b+c \le (a^3+b^3+c^3)^{1/3} 3^{2/3},$$
and next with $x = a^2$, $y = b^2$, $z = c^2$ and $p = 3/2$ to get
$$a^2+b^2+c^2 \le (a^3+b^3+c^3)^{2/3} 1/3.$$
Equality holds in both iff $a = b = c$. In any event, forming their product, we get
$$(a+b+c)(a^2+b^2+c^2) \le 3(a^3+b^3+c^3).$$

A standard ploy used to deal with inequalities involving the side lengths $a$, $b$, $c$ of a triangle is to reformulate them in terms of the positive variables
$$x = a + b - c, \quad y = b + c - a, \quad z = c + a - b;$$
so that
$$a = \frac{z+x}{2}, \quad b = \frac{x+y}{2}, \quad c = \frac{y+z}{2},$$
whence $a + b + c = x + y + z$,
$$a^2 + b^2 + c^2 = \frac{1}{2}(x^2 + y^2 + z^2 + xy + yz + zx) = \frac{1}{2}(\sum x^2 + \sum xy),$$
$$(a+b+c)(a^2+b^2+c^2) = \frac{1}{2}\left(\sum x^3 + 2\sum xy(x+y) + 3xyz\right)$$
and
$$a^3 + b^3 + c^3 = \frac{1}{8}(2\sum x^3 + 3\sum xy(x+y)).$$
To establish the left inequality, notice that
$$\begin{aligned} (a+b+c)(a^2+b^2+c^2) - 2(a^3+b^3+c^3) \\ = \frac{1}{2}\left(\sum x^3 + 2\sum xy(x+y) + 3xyz\right) - \frac{1}{4}\left(2\sum x^3 + 3\sum xy(x+y)\right) \\ = \frac{1}{4}\left(\sum xy(x+y) + 6xyz\right), \end{aligned}$$
which is clearly positive. Hence the left inequality follows.
Also,
$$\begin{aligned} 3(a^3 + b^3 + c^3) - (a+b+c)(a^2 + b^2 + c^2) \\ = \frac{1}{8}(2(x^3 + y^3 + z^3) - 12xyz + x^2(y+z) + y^2(z+x) + z^2(x+y)) \\ = \frac{1}{8}(x^3 + y^3 + z^3 - 12xyz + (x^2 + y^2 + z^2)(x+y+z)) \\ = \frac{1}{8}((x^3 + y^3 + z^3 - 3xyz) + ((x^2 + y^2 + z^2)(x+y+z) - 9xyz)) \\ \ge 0, \end{aligned}$$
by three applications of the AM-GM inequality, with equality iff $x = y = z$.
In other words, the second inequality holds.

To establish the inequality on the left, multiply the three inequalities $a < b+c$, $b < c+a$, and $c < a+b$ by $a^2$, $b^2$, and $c^2$, respectively. This results in
$$a^3 < a^2(b+c), \quad b^3 < b^2(c+a), \quad c^3 < c^2(a+b).$$
Adding these together and then adding $a^3 + b^3 + c^3$ to both sides of the resulting inequality gives the desired inequality. This is essentially the same proof as in Solution 1.
The inequality on the right holds for any triple of positive numbers $a$, $b$, $c$. To establish this, similar to Solution 2, but not switching to $x$, $y$, $z$, we multiply out the middle term and subtract it from $3(a^3 + b^3 + c^3)$, to see that we are left to show that
$$2(a^3 + b^3 + c^3) - ab^2 - a^2b - bc^2 - b^2c - ca^2 - c^2a \ge 0. \quad (9)$$
But
$$\begin{aligned} & 2(a^3 + b^3 + c^3) - ab^2 - a^2b - bc^2 - b^2c - ca^2 - c^2a \\ &= (a^3 + b^3 - a^2b - b^2a) + (b^3 + c^3 - b^2c - c^2b) + (c^3 + a^3 - c^2a - a^2c) \\ &= (a^2 - b^2)(a - b) + (b^2 - c^2)(b - c) + (c^2 - a^2)(c - a) \\ &= (a+b)(a-b)^2 + (b+c)(b-c)^2 + (c+a)(c-a)^2 \ge 0, \end{aligned}$$
since $a$, $b$, $c$ are positive, with equality iff $a = b = c$. Hence the second inequality holds.