Let $x, y$ be real numbers whose sum is 1. Determine the maximum value of $x y^{4}+x^{4} y$.

 untranslated portion:
Az $x, y$ valós számok összege 1. Határozzuk meg $x y^{4}+x^{4} y$ legnagyobb értékét.

 translated portion:
Let $x, y$ be real numbers whose sum is 1. Determine the maximum value of $x y^{4}+x^{4} y$.
I. solution. The expression under investigation is symmetric, so it can be written as a polynomial in terms of $x+y(=1)$ and $x y=t$. Indeed:

$$
A(x, y)=x y^{4}+x^{4} y=x y\left(x^{3}+y^{3}\right)=x y\left[(x+y)\left(x^{2}-x y+y^{2}\right)\right]
$$

Since $x^{2}-x y+y^{2}=(x+y)^{2}-3 x y$, we have

$$
A(x, y)=x y\left[(x+y)\left((x+y)^{2}-3 x y\right)\right]=x y(1-3 x y)
$$

If $x+y=1$, then the value set of the product $x y=t$ is the interval ] $-\infty ; 1 / 4]$, for the discriminant of the quadratic equation derived from the system

$$
\left.\begin{array}{rl}
x+y & =1 \\
x y & =t
\end{array}\right\}
$$

is $D=1-4 t$, and the system has real solutions if and only if $t \leq \frac{1}{4}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_41f29154fe85e7edcd1dg-1.jpg?height=485&width=631&top_left_y=760&top_left_x=736)

# Figure 1

Thus, under the condition $x+y=1$, the value set of the two-variable polynomial $A(x, y)$ on the interval $I=]-\infty ; 1 / 4]$ coincides with the value set of the function $f(t)=t(1-3 t)$ (Figure 1). The function $f$ extended to the set of real numbers attains its absolute maximum at $t=\frac{1}{6}$, and this value is $f\left(\frac{1}{6}\right)=\frac{1}{12}$. Since $\frac{1}{6} \in I$, this is the maximum of the function $f(t)$ defined on the interval $I$, and thus also the maximum value of $A(x, y)$ under the condition $x+y=1$.

Csaba Pálinkás (Szolnok, Verseghy F. Gymnasium, 9th grade)

Remarks. 1. Most of the solvers wrote the expression $x y^{4}+x^{4} y$ in the form $-3\left(x y-\frac{1}{6}\right)^{2}+\frac{1}{12}$ using the condition $x+y=1$. In more than 100 solutions, after this, something appeared without any further explanation, stating that the sought maximum value is $\frac{1}{12}$. This is only true if the first term of the sum, $-3\left(x y-\frac{1}{6}\right)^{2}$, attains the value 0, i.e., there are $x$ and $y$ whose sum is 1 and whose product is $\frac{1}{6}$. This can be seen from the solution, although these values were not calculated. Solving the system

$$
\left.\begin{array}{rl}
x+y & =1 \\
x y & =\frac{1}{6}
\end{array}\right\}
$$

immediately yields the values of the variables for which $A(x, y)$ is maximal: $\{x, y\}=$ $\left\{\frac{3-\sqrt{3}}{6}, \frac{3+\sqrt{3}}{6}\right\}$

2. According to the result of the problem, if $x+y=1$, then $x y^{4}+x^{4} y \leq \frac{1}{6}$. In this form, we can see an example of the rare phenomenon where in a symmetric inequality, the maximum is not attained when the variables are equal.

II. solution. We solve the problem using the application of differential calculus. The originally two-variable function can be written as a one-variable polynomial by the substitution $y=1-x$:

$$
f(x)=x(1-x)^{4}+x^{4}(1-x)
$$

Let's write down the derivative of the function, and after performing the exponentiation, rearrange the polynomial:

$$
f^{\prime}(x)=(1-x)^{4}-4 x(1-x)^{3}+4 x^{3}(1-x)-x^{4}=-12 x^{3}+18 x^{2}-8 x+1
$$

Since the original two-variable polynomial is symmetric, it follows that $f(x)=f(1-x)$. For the derivative, this implies $f^{\prime}(x)=$ $-f^{\prime}(1-x)$, and thus $f^{\prime}\left(\frac{1}{2}\right)=-f^{\prime}\left(\frac{1}{2}\right)$, which leads to $f^{\prime}\left(\frac{1}{2}\right)=0$. Based on this, the derivative can be factored as:

$$
f^{\prime}(x)=\left(x-\frac{1}{2}\right)\left(-12 x^{2}+12 x-2\right)
$$

The quadratic factor has two real roots, $x_{1}=\frac{3-\sqrt{3}}{6}$ and $x_{2}=\frac{3+\sqrt{3}}{6}$. Note that since $x_{1}+x_{2}=1$, we have $f\left(x_{1}\right)=f\left(x_{2}\right)$.

Since the cubic $f^{\prime}(x)$ has three distinct real roots and its leading coefficient is negative, its graph can be sketched: (Figure 2). The sign of the derivative can be read from the graph, and thus it is also clear that $f$ has two local maxima: at $x_{1}$ and $x_{2}$. We have already seen that these maxima are equal, and since $f$ is strictly increasing if $xx_{2}$, this common value, $f\left(x_{1}\right)=f\left(x_{2}\right)=\frac{1}{12}$, is the absolute maximum of the function.

## Figure 3

Péter Szilágyi (Debrecen, Kossuth L. Gymnasium, 10th grade)

Remark. The above examination also reveals that $f$ has a local minimum at $x=\frac{1}{2}$; its value is $f\left(\frac{1}{2}\right)=$ $\frac{1}{16}$. The graph of the function $f$ is shown in Figure 3.