Calculate the maximum value of $x^3y + y^3z + z^3x$ where $x + y + z = 4$ and $x, y, z \ge 0$.

Question

Given non-negatives $x, y, z$ such that $x + y + z = 4$. Calculate the maximum value of $$\large x^3y + y^3z + z^3x$$

As an assumption, the maximum value is $27$, occured when $(x, y, z) = (0, 1, 3)$.

I have a guess about a working-in-process prove. Let $y$ be the median of $x, y, z$.

$$\iff (zx - yz)(y^2 - z^2) \ge 0 \iff y^2zx - y^3z - z^3x + yz^3 \ge 0$$

$$\iff x^3y + y^3z + z^3x \le x^3y + y^2zx + yz^3 = y(x^3 + xyz + z^3)$$

And further down the line is what I haven't accomplished yet.

Answer 1

Now, use AM-GM: $$y(x^3+xyz+z^3)\leq y(x+z)^3=27y\left(\frac{x+z}{3}\right)^3\leq27\left(\frac{y+3\cdot\frac{x+z}{3}}{4}\right)^4=27.$$

Answer 2

I will give you a hint for the generalized problem:

$$f(x,y,z) = x^my^n+y^mz^n+z^mx^n\leq\dfrac{m^mn^n}{(m+n)^{m+n}}$$ for non-negative reals satisfying $x+y+z = 1$ and $m\geq n\geq 0.$ The simplest proof starts out by assuming $z = \max\{x,y,z\}$ and shows: $$f(x,y,z)\leq f(0,x+y,z) = f(0,1-z,z),$$ and the RHS can be dealt with by AM-GM. This is a direct generalization but the proofs in the link on the comment in your post apply as well. The equality would be attained at any cyclic permutation of : $$\left(0,\frac{n}{m+n}, \frac{m}{m+n}\right)$$

For your problem, you need to normalize first.

Answer 3

I like the following way.

Let $\{x,y,z\}=\{a,b,c\}$, where $a\geq b\geq c$.

Thus, by Rearrangement and AM-GM we obtain: $$x^3y+y^3z+z^3x=x^2\cdot xy+y^2\cdot yz+z^2\cdot zx\leq a^2\cdot ab+b^2\cdot ac+c^2\cdot bc=$$ $$=b(a^3+c^3+abc)\leq b(a+c)^3=27\cdot b\left(\frac{a+c}{3}\right)^3\leq27\left(\frac{b+3\cdot\frac{a+c}{3}}{4}\right)^4=27.$$

Answer 4

Assume $x\equiv \max\{x, y, z\}$, I found "$=$" when $x- 3y= z= 0$, let's prove the homogeneous inequality $$27(\sum\limits_{cyc} x)^{4}- 256(\sum\limits_{cyc} x^{3}y)= z\left ( 148\left ( xz(x- z)+ y^{2}(x- y) \right )+ 108(yz^{2}+ x^{3})+ 324xy(x+ z)+ 27z^{3}+ 14x^{2}z+ 162y^{2}z+ 176xy^{2} \right )+ (x- 3y)^{2}(27x^{2}+ 14xy+ 3y^{2})\geqq 0$$

Given $n$ non-negative numbers $x_{1}, x_{2}, etc, x_{n}$ so that $x_{1}+ x_{2}+ etc+ x_{n}= 1$ but $((n> 3)\,or\,(n= 3, mk\neq 1))$ and $m, k> 0$. Prove that $$x_{1}^{m}x_{2}^{k}+ x_{2}^{m}x_{3}^{k}+ etc+ x_{n}^{m}x_{1}^{k}\leqq \frac{m^{m}k^{k}}{(m+ k)^{m+ k}}$$