Maximum of $xy^4+x^4y$ given $x+y=1$

Question

Question:Maximise $x^4y+y^4x$ given $x+y=1$ and $x,y>0$

This is probably a very simple question but i noticed that we get a minimum when $x=y=0.5$.Of course calculus seems to be a reliable option but i think the expression gets very ugly.

I couldn't proceed much.

WLOG $x\ge y$.By Chebyshov $$2(x^4y+y^4x)\le (x+y)(x^4+y^4)=x^4+y^4=1-4x^3y-4y^3x-6x^2y^2$$

Also $x=\sin^2 t,y=\cos^2 t$ doesn't help.

I don't know how to approach the problem. I would be very thankful if someone could give me a hint. I am preferably looking for a solution using some basic inequalities,

Answer 1

Rewrite as \begin{eqnarray*} x^4y+y^4x&=&xy(x+y)(x^2-xy+y^2)=xy(x+y)((x+y)^2-3xy)=p(1-3p) \\ &=&\frac{1}{12}-3(p-\frac{1}{6})^2. \end{eqnarray*}

Answer 2

Notice that $$x^4y+xy^4=xy(x^3+y^3)=xy(x+y)(x^2-xy+y^2)=xy(x+y)((x+y)^2-3xy)=xy(1-3xy)$$ Now, by AM-GM $$\frac{3xy+(1-3xy)}{2}\geqslant \sqrt{3xy(1-3xy)}\iff \frac{1}{12}\geqslant xy(1-3xy)$$

Answer 3

Let $x=\dfrac12+t$ which implies $y=\dfrac12-t$. The objective function is

$$\left(\frac12+t\right)\left(\frac12-t\right)^4+\left(\frac12+t\right)^4\left(\frac12-t\right)=-3t^4+\frac{t^2}2+\frac1{16}$$ which is quadratic in $t^2$.

By completing the square,

$$-3\left(t^2-\frac1{12}\right)^2+\frac1{12}\le\frac1{12}$$ and this is tight.

Answer 4

$$x^4y+xy^4=xy(x^3+y^3)$$

$$=xy((x+y)^3-3xy(x+y))$$

$$=xy(1-3xy)$$

$$=\dfrac{12xy-(6xy)^2}{12}$$

$$=\dfrac{1-(6xy-1)^2}{12}\le\dfrac1{12}$$

which occurs if $1=6xy=6x(1-x)\iff6x^2-6x+1=0\implies x=?$