Find $\max xy^3z^7$ where $x+y+z=1$

As a brief introduction, 'problem of the day' posts will be about mathematical thinking (problem-solving and proofs). The questions and solutions are extracted from online sources. I will add alternative solutions and mention something interesting about the problems if I can.

Today's problem:

Find the maximum of $xy^3z^7$ with non-negative real numbers in the plane $x+y+z=1$.

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We make use of AM-GM inequality: $$1=x+y+z=x+\frac{1}{3}y+\frac{1}{3}y+\frac{1}{3}y+\frac{1}{7}z+\cdots+\frac{1}{7}z\geq 11 \sqrt[11]{x(\frac{1}{3}y)^3(\frac{1}{7}z)^7}.$$ To obtain the maximum, we want to find $x,y,z$ for the equality case. Namely, $x=\dfrac{1}{3}y=\dfrac{1}{7}z=\dfrac{1}{11}$. We thus have the maximum $xy^3z^7=\dfrac{3^3 7^7}{11^{11}}$.

Alternatively, we can use lagrange multiplier since this is an optimisation problem with a constraint. Let $F(x,y,z,\lambda)=xy^3z^7-\lambda(x+y+z-1)$. Then, we have a system of equations $$\begin{cases}F_x=y^3z^7-\lambda=0\\F_y=3xy^2z^7-\lambda=0\\F_z=7xy^3z^6-\lambda=0\\ x+y+z=1.\end{cases}$$ From the first three equations, we know $y=3x,z=7x$. Therefore, substituting them into the last equation, we have $x+3x+7x=1$ and thus $x=\dfrac{1}{11},y=\dfrac{3}{11},z=\dfrac{7}{11}$.