I will show a bit more, namely that the cubic polynomial in question has three real roots, the smallest of which is less than $2/5$. This is not tight: further improvements can be squeezed out of the method.

Let $u$ and $v$ be as in Iosif Pinelis's response. Then the cubic polynomial in question equals
$$p(x):=10 x^3 - 30 x^2 + (30 - 2 u) x + (2 u - v - 10),$$
and its discriminant $20(16 u^3 - 135 v^2)$ is nonnegative by [this answer of Fedor Petrov][1]. So $p(x)$ has three real roots (counted with multiplicity). Let us assume that all these roots are at least $2/5$. Then the three roots of
$$p(2/5+x)=10x^3-18x^2+\left(\frac{54}{5}-2u\right)x+\left(\frac{6u}{5}-v-\frac{54}{25}\right)$$
are nonnegative, hence in particular
$$\frac{54}{5}-2u\geq 0\geq\frac{6u}{5}-v-\frac{54}{25}.$$
It follows that $u\leq 27/5$ and $u-v\leq 54/25$. However, my response [for this other MO question][2] shows that $|u-v-3|\leq(4/5)^{3/2}$, which contradicts $u-v\leq 54/25$.


  [1]: https://mathoverflow.net/a/479088/36721
  [2]: https://mathoverflow.net/q/479107/