Irish Mathematical Olympiad

a.
Let $x'$, $y'$, $z'$ and $S'$ be obtained from $x$, $y$, $z$ and $S$ by replacing $a$, $b$, $c$ by $a + t$, $b + t$, $c + t$, respectively. So
$$x' = (a + t) + (b + t) + (c + t) = x + 3t,$$
$$\begin{aligned} y' &= (a + t)^2 + (b + t)^2 + (c + t)^2 = (a^2 + b^2 + c^2) + 2(a + b + c)t + 3t^2 \\ &= y + 2xt + 3t^2, \end{aligned}$$
$$\begin{aligned} z' &= (a + t)^3 + (b + t)^3 + (c + t)^3 \\ &= (a^3 + b^3 + c^3) + 3(a^2 + b^2 + c^2)t + 3(a + b + c)t^2 + 3t^3 \\ &= z + 3yt + 3xt^2 + 3t^3. \end{aligned} \text{ Hence}$$
$$\begin{aligned} S' &= 2x'^3 - 9x'y' + 9z' \\ &= 2(x + 3t)^3 - 9(x + 3t)(y + 2xt + 3t^2) + 9(z + 3yt + 3xt^2 + 3t^3) \\ &= 2x^3 - 9xy + 9z = S \quad (\text{All other terms cancel.}) \end{aligned}$$

b.
If we let $T = 3y - x^2$ and $T' = 3y' - x'^2$, we obtain
$$\begin{aligned} T &= 3(a^2 + b^2 + c^2) - (a + b + c)^2 = 2(a^2 + b^2 + c^2 - ab - ac - bc) \\ &= (a - b)^2 + (b - c)^2 + (c - a)^2 \\ &= (a' - b')^2 + (b' - c')^2 + (c' - a')^2 = T'. \end{aligned}$$

Version A. We take $t = -c$ and so it suffices to prove that $T^3 - 2S^2 \ge 0$ in the case $c = 0$, where we get $T = 3(a^2 + b^2) - (a + b)^2 = 2(a^2 - ab + b^2)$ and
$$\begin{aligned} S &= 2(a + b)^3 - 9(a + b)(a^2 + b^2) + 9(a^3 + b^3) \\ &= 2a^3 - 3a^2b - 3ab^2 + 2b^3 \\ &= 2(a^3 + b^3) - 3ab(a + b) \\ &= (a + b)(2(a^2 - ab + b^2) - 3ab). \end{aligned} \quad \text{Now}$$
$$\begin{aligned} T^3 - 2S^2 &= 8(a^2 - ab + b^2)^3 - 2(a + b)^2(2(a^2 - ab + b^2) - 3ab)^2 \\ &= 8(a^2 - ab + b^2)^3 - 8(a + b)^2(a^2 - ab + b^2)^2 \\ &\quad + 24ab(a + b)^2(a^2 - ab + b^2) - 18a^2b^2(a + b)^2 \\ &= -24ab(a^2 - ab + b^2)^2 + 24ab(a + b)^2(a^2 - ab + b^2) - 18a^2b^2(a + b)^2 \\ &= 24ab(a^2 - ab + b^2)((a + b)^2 - (a^2 - ab + b^2)) - 18a^2b^2(a + b)^2 \\ &= 72a^2b^2(a^2 - ab + b^2) - 18a^2b^2(a^2 + 2ab + b^2) \\ &= 6a^2b^2(12(a^2 + b^2) - 12ab - 3(a^2 + b^2) - 6ab) \\ &= 54a^2b^2(a - b)^2 \ge 0. \quad \text{This completes the proof.} \end{aligned}$$

Version B. By choosing $t = -(a + b + c)/3$ we achieve that $x$ vanishes. In this case, $T = 3y$ and $S = 9z$ and $c = -(a + b)$. We obtain
$$\begin{aligned} y &= a^2 + b^2 + (a + b)^2 = 2(a^2 + ab + b^2) \\ z &= a^3 + b^3 - (a + b)^3 = -3ab(a + b) \end{aligned}$$

and so $T = 6(a^2 + ab + b^2)$ and $S = -27ab(a + b)$. Hence,
$$(T^3 - 2S^2)/54 = 4(a^2 + ab + b^2)^3 - 27(ab(a + b))^2.$$

$$(T^3 - 2S^2)/54 = 4(a^2 + a + 1)^3 - 27(a(a + 1))^2.$$
As this vanishes for $a = 1$, we rewrite it in terms of multiples of $a - 1$. Because $a(a + 1) = (a - 1)(a + 2) + 2$ and $a^2 + a + 1 = (a - 1)(a + 2) + 3$, we see that the above vanishes for $a = -2$, too. Abbreviating $P = (a - 1)(a + 2)$, we get
$$\begin{aligned} (T^3 - 2S^2)/54 &= 4(P + 3)^3 - 27(P + 2)^2 \\ &= 4(P^3 + 9P^2 + 27P + 27) - 27(P^2 + 4P + 4) \\ &= 4P^3 + 9P^2 = P^2(4P + 9) \\ &= (a - 1)^2(a + 2)^2(4(a - 1)(a + 2) + 9) \\ &= (a - 1)^2(a + 2)^2(4a^2 + 4a + 1) \\ &= (a - 1)^2(a + 2)^2(2a + 1)^2 \geq 0. \end{aligned}$$

We now set $P = a^2 + ab + b^2$ and $Q = 3a^2$. We have $27(ab(a + b))^2 = 27a^2(R - a^2)^2 = 3a^2(3R - 3a^2)^2 = Q(3R - Q)^2$, and so
$$(T^3 - 2S^2)/54 = 4R^3 - Q(3R - Q)^2.$$

This vanishes for $Q = R$, so we write $3R - Q = 2R + (R - Q)$ and simplify:
$$\begin{align*} 4R^3 - Q(2R + (R - Q))^2 &= 4R^3 - 4R^2Q - 4QR(R - Q) - Q(R - Q)^2 \\ &= 4R^2(R - Q) - (4QR + Q(R - Q))(R - Q) \\ &= (R - Q)(4R^2 - 4QR - Q(R - Q)) \\ &= (R - Q)(4R(R - Q) - Q(R - Q)) \\ &= (R - Q)^2(4R - Q) = (R - Q)^2(a^2 + 4ab + 4b^2) \\ &= (R - Q)^2(a + 2b)^2 \ge 0. \end{align*}$$

Remark. We may trace back our steps and write $a/b$ for $a$ and multiply by $b^6$. This translates $(a - 1)^2(a + 2)^2(2a + 1)^2$ into $(a - b)^2(a + 2b)^2(2a + b)^2$. Writing $-c$ for $a + b$ in the last two brackets, we get $(a - b)^2(b - c)^2(a - c)^2$, which is the discriminant of the cubic $(X - a)(X - b)(X - c)$.

a.
For every integer $k \ge 0$ we define a function
$$x_k(t) = (a + t)^k + (b + t)^k + (c + t)^k$$
which is defined for all real numbers $t$. We also let
$$S(t) = 2x_{1}(t)^{3} - 9x_{1}(t)x_{2}(t) + 9x_{3}(t) \quad \text{and} \quad T(t) = 3x_{2}(t) - x_{1}(t)^{2},$$
so that $S = S(0)$ and $T = T(0)$. Noting that $\frac{dx_k}{dt} = kx_{k-1}$ for $k \ge 1$, we easily obtain $\frac{dS}{dt} = 0$ and $\frac{dT}{dt} = 0$. In other words, $S(t)$ and $T(t)$ are constant functions. This establishes part (a) and also shows that $T(t)^3 - 2S(t)^2$ does not depend on $t$.

b.
We let $m = (a + b + c)/3$ and obtain $x_1(-m) = 0$, hence
$$T = T(-m) = 3x_2(-m) = 3((a - m)^2 + (b - m)^2 + (c - m)^2)$$
and
$$\begin{align*} S &= S(-m) = 9x_3(-m) \\ &= 9((a - m)^3 + (b - m)^3 + (c - m)^3) \\ &= 27(a - m)(b - m)(c - m), \end{align*}$$

since $u + v + w$ is a factor of $u^3 + v^3 + w^3 - 3uvw$. The inequality $T^3 \ge 2S^2$ is now seen to be equivalent to
$$((a - m)^2 + (b - m)^2 + (c - m)^2)^3 \ge 54(a - m)^2(b - m)^2(c - m)^2$$
which is a consequence of the following lemma.

Lemma. If $p$, $q$, $r$ are real numbers such that $p + q + r = 0$, then
$$54p^2q^2r^2 \leq (p^2 + q^2 + r^2)^3.$$
Proof. This is clear if $pqr = 0$. Suppose $pqr \neq 0$, then at least one of $pq$, $pr$, $qr$ is negative. Say $pq < 0$, so that $p^2 + q^2 + r^2 = (p + q)^2 - 2pq + r^2 = 2(r^2 + |pq|)$. The AM-GM inequality gives $\frac{|pq|}{2} \cdot \frac{|pq|}{2} \cdot r^2 \leq \left(\frac{r^2 + |pq|}{3}\right)^3$ and so
$$54p^2q^2r^2 = 2^3 \cdot 3^3 \cdot \frac{|pq|}{2} \cdot \frac{|pq|}{2} \cdot r^2 \leq 8(r^2 + |pq|)^3 = (p^2 + q^2 + r^2)^3.$$