China Hong Kong Mathematical Olympiad

The greatest $K$ is $16$.

We first prove the inequality when $K = 16$. Note that
$$\begin{aligned} u^2 - 4vw &= u^2 + 2vw - 6vw \\ &\ge u^2 + 2vw - 3(v^2 + w^2) \\ &= u^2 + (v+w)^2 - 4(v^2 + w^2) \\ &\ge 2u(v+w) - 4(v^2 + w^2) \\ &= 2(uw - 2v^2) + (uv - 2w^2). \end{aligned}$$
If $uw - 2v^2$ and $uv - 2w^2$ have different signs (possibly equal to $0$), we have
$$(u^2 - 4vw)^2 \ge 0 \ge K(2v^2 - uw)(2w^2 - uv).$$
The equality case will be handled below.
If both $uw - 2v^2$ and $uv - 2w^2$ are negative, then $(uw)(uv) < (2v^2)(2w^2)$, which yields the contradiction $u^2 < 4vw$.
Therefore, we may assume both $uw - 2v^2$ and $uv - 2w^2$ are positive. Therefore, we obtain
$$(u^2 - 4vw)^2 \ge 4((uw - 2v^2) + (uv - 2w^2))^2 \ge 16(uw - 2v^2)(uv - 2w^2).$$
For the equality to hold, we must have $v = w$ and $u = v + w$. This implies
$$u^2 = (2v)^2 = 4v^2 = 4vw,$$
contradicting $u^2 > 4vw$. Therefore, equality cannot hold, and the inequality is proven.

Next, we show $K = 16$ is the largest possible. Consider $v = w = 1$ and $u = 2 + \varepsilon$ for a sufficiently small positive $\varepsilon$. The relation $u^2 > 4vw$ holds. Also, we have
$$(u^2 - 4vw)^2 = (4 + 4\varepsilon + \varepsilon^2 - 4)^2 = \varepsilon^2(4 + \varepsilon)^2$$
and
$$K(2v^2 - uw)(2w^2 - uv) = K(2 - 2 - \varepsilon)^2 = K\varepsilon^2.$$
The inequality holds if and only if $(4 + \varepsilon)^2 > K$. When $\varepsilon$ approaches $0$, the left-hand side approaches $16$. Therefore, $K \le 16$. This completes the proof.