# $x,y,z \geqslant 0$, $x+y^2+z^3=1$, prove $x^2y+y^2z+z^2x < \frac12$

$x,y,z \geqslant 0$, $x+y^2+z^3=1$, prove $$x^2y+y^2z+z^2x < \frac12$$

This inequality has been verified by Mathematica. $\frac12$ is not the best bound. I try to do AM-GM for this one but not yet success. The condition $x+y^2+z^3$ is very weird.

• When $x=0.534,y=0.5$ and $z=0.6$ (exact decimal values) one has $x^2y+y^2z+z^2x=0.484818$, so 0.5 is close to the best constant. – Ewan Delanoy May 21 '16 at 18:19
• Mathematica gives the maximum value of $x^2y+y^2z+z^2x < \frac12$ as constrained to be $0.48562209920309984177$ for $x=0.55811253$, $y=0.49841201$, $z=0.57837131$. – Steve Kass May 26 '16 at 3:23
• I think, this method will be interesting too. – Yuri Negometyanov Apr 6 '17 at 11:25

## 7 Answers

The standard way of solving the problem on a conditional extremum is the method of Lagrange multipliers, which reduces it to a system of equations.

The greatest value of function $$f(x,y,z,\lambda) = x^2y+y^2z+z^2x+\lambda(x+y^2+z^3-1)$$ on the interval $$x,y,z\in[0,1]$$ is reached or at its edges, or in one of the points with zero partial derivatives $$\begin{cases} f'_\lambda = x + y^2 + z^3 - 1 = 0\\ f'_x = z^2 + 2xy + \lambda = 0\\ f'_y = x^2 + 2yz + 2\lambda y = 0\\ f'_z = y^2 + 2zx + 3\lambda z^2 = 0. \end{cases}$$ After the excluding of parameter $\lambda$ get the system $$\begin{cases} x + y^2 + z^3 - 1 = 0\\ x^2 + 2yz = 2y(z^2 + 2xy)\\ y^2 + 2zx = 3z^2(z^2 + 2xy) \end{cases}$$ with positive solutions $$\genfrac{[}{.}{0}{0}{x\approx 0.16367,\quad y\approx 0.761982,\quad z\approx 0.634724,\quad f\approx 0.454882} {x\approx 0.558113,\quad y\approx 0.498412,\quad z\approx 0.578371,\quad f\approx 0.485622}.$$ The edges of the field are achieved when $x = 0$, $y = 0$ or $z = 0$.

Substituting value $x = 0$ to the expressions for the partial derivatives, we have $$\begin{cases} x=0\\ y^2+z^3 = 1\\ 2yz+2\lambda y = 0\\ y^2+3\lambda z^2 = 0 \end{cases}$$ with solutions $$\genfrac{[}{.}{0}{} {x=0,\quad y=\sqrt{0.75},\quad z=\sqrt{0.25}\approx 0.629991,\quad f\approx 0.47247} {x=0,\quad y=0,\quad z=1,\quad f=0}$$

Substituting value $y = 0$ to the expressions for the partial derivatives, we have $$\begin{cases} y=0\\ x+z^3=1\\ z^2+\lambda = 0\\ 2zx+3\lambda z^2 = 0 \end{cases}$$ with solution $$x=0.6,\quad y=0,\quad z=\sqrt{0.4}\approx 0.736806,\quad f\approx 0.32573.$$

Substituting value $z = 0$ to the expressions for the partial derivatives, we have $$\begin{cases} z=0\\ x+y^2=1\\ 2xy+\lambda = 0\\ x^2+2\lambda y = 0 \end{cases}$$ with solutions $$\genfrac{[}{.}{0}{} {x=0.8,\quad y=\sqrt{0.2}\approx 0.447214,\quad z=0,\quad f\approx 0.286217} {x=0,\quad y=0,\quad z=1,\quad f=0}$$

Account, that the values of function at the vertices of the area (unit parallelepiped) are zero.

So required maximal value approximately equal to $0.485622$. Given the accuracy of calculations it guarantees the inequality $$\boxed{x^2y+y^2z+z^2x<1/2.}$$

System Resolving (updated 28.08.16)

Set condition allows us to reduce the problem to finding unconditional extremes of $$f(y,z)=(1-y^2-z^3)^2y+y^2z+z^2(1-y^2-z^3).$$ Necessary optimality conditions in the field have the form: $$\begin{cases} f'_y = (1-y^2-z^3)^2-4y^2(1-y^2-z^3)+2yz-2yz^2 = 0\\ f'_z = -6yz^2(1-y^2-z^3)+y^2+2z(1-y^2-z^3)-3z^4 = 0, \end{cases}$$ or $$\begin{cases} 5y^4+y^2(6z^3-6)+y(2z-2z^2)+(z^3-1)^2 = 0\\ 6y^3z^2+y^2(1-2z)+6y(z^3-1)z^2-5z^4+2z = 0. \end{cases}$$

If to consider the coefficient of the highest power of $y$ as denominator in equation, and the remaining coefficients - numerators, we get the above equations. Then we can subtract the second equation factor $y$ from the first and repeat subtraction with factor $1$, obtaining the system

$$\begin{cases} C_{2,2}(z)y^2 + C_{2,1}(z)y + C_{2,0}(z) = 0\\ 6y^3z^2+y^2(1-2z)+6y(z^3-1)z^2-5z^4+2z = 0, \end{cases}$$ where $$C_{2,2}(z) = 36z^7-36z^4+20z^2-20z+5,$$ $$C_{2,1}(z) = 18z^6+102z^5-30z^2,$$ $$C_{2,0}(z) = 36z^{10}-72z^7+50z^5+11z^4-20z^2+10z.$$

Thus, the order of the first equation in $y$ reduced from fourth to second. Likewise, lowering the order of the second equation by a first equation, we obtain

$$\begin{cases} C_{2,2}(z)y^2 + C_{2,1}(z)y + C_{2,0}(z) = 0\\ D_{2,2}(z)y^2 + D_{2,1}(z)y + D_{2,0}(z) = 0,\\ \end{cases}\qquad(1)$$ where $$D_{2,2}(z) = 180z^8+576z^7-72z^5-144z^4+40z^3-60z^2+30z-5,$$ $$D_{2,1}(z) = 180z^7-30z^6-30z^5-60z^3+30z^2,$$ $$D_{2,0}(z) = 180z^{11}-252z^8+100z^6-28z^5+25z^4-40z^3+40z^2-10z.$$

The system $(1)$ is a linear in the unknowns $y^2$ and $y$, so $$y^2=\dfrac{\Delta_2(z)}{\Delta_0(z)},\quad y=\dfrac{\Delta_1(z)}{\Delta_0(z)},\qquad(2)$$ where $$\Delta_0(z) = C_{2,2}(z)D_{2,1}(z) - C_{2,1}(z)D_{2,2}(z),$$ $$\Delta_2(z) = -C_{2,0}(z)D_{2,1}(z) + C_{2,1}(z)D_{2,0}(z),$$ $$\Delta_1(z) = -C_{2,2}(z)D_{2,0}(z) + C_{2,0}(z)D_{2,2}(z),$$ or $$\Delta_0(z) = 90z^{10}-828z^9-1662z^8-144z^7+396z^6+1028z^5-200z^4+180z^3-220z^2+10z,$$ $$\Delta_2(z) = -90z^{13}+540z^{12}+30z^{11}+234z^{10}-864z^9-290z^8+256z^7+74z^6+220z^5-160z^4+100z^3-50z^2,$$ $$\Delta_1(z) = 576z^{13}-1296z^{10}+90z^9+815z^8+444z^7-5z^6-580z^5+126z^4+10z^3+80z^2-30z-5.$$

From $(2)$ for $\Delta_0(z)\not=0$ should be $$\Delta_1^2(z) - \Delta_2(z)\Delta_0(z) = 0,$$ or $$331776z^{26}-1484892z^{23}-19440z^{22}+1233720z^{21}+3079404z^{20}+195732z^{19}-3189924z^{18}-3109428z^{17}+368233z^{16}+2734116z^{15}+1243978z^{14}-741000z^{13}-805907z^{12}+75696z^{11}+164040z^{10}+82560z^9-172194z^8+53440z^7-10290z^6+32440z^5-7460z^4-4400z^3+100z^2+300z+25 = 0.$$

The coefficients are particially calculated using the Mathcad package, and $\mathcal{polyroots}()$ function is also used, which calculates all the roots of the polynomial by the "accompanying matrix" method.

Calculating values $y$ with $(2)$ and $x,f$ by the formula $$x = 1-y^2-z^3,\quad f=xy^2+yz^2+zx^2$$ and checking them by substituting in the original system, we obtain the following stationary points with the positive coordinates: $$(x,y,z,f)\in\left[\genfrac{}{}{0pt}{0} {(0.5581125,\ 0.4984120,\ 0.5783713,\ 0.4856221)} {(0.1636702,\ 0.7619816,\ 0.6347238,\ 0.4548812)} \right.$$

• Less time consuming way to set out as a separate answer. – Yuri Negometyanov May 27 '16 at 12:43
• The problem with this is that the equations were solved using some solver (approximations) and so cant represent exact solution – ibnAbu May 27 '16 at 18:49
• Worst there is no proof that the approximate solutions give global maximum. – ibnAbu May 27 '16 at 18:50
• @stalker2133 Approximation seems a single way to get exact maximum, because we've got a hard task. "Classic" (clear) proof is shown there in another my answer, it was in my link. But never note it.))) – Yuri Negometyanov May 27 '16 at 19:44
• I checked the link but I wonder why we are using determinant to tell about solutions of non- linear equations. – ibnAbu May 27 '16 at 21:16

This answer is incomplete.

Let $A=x+y^2$, $B=y^2+z^3$ and $C=z^3+x$.

Claim: $x^2y \leq \dfrac{A}{2}x^{3/2}$, $y^2z \leq \dfrac{B^{2/3}}{4^{2/3}}y^{4/3}$, $z^2x \leq \dfrac{C^{4/3}}{4^{2/3}}x^{1/3}$.

Assuming the claim, we note that $A+B+C=2$ and we need to prove that the given sum is less than $\dfrac{A+B+C}{4}$.

• Isn't a+b+c=2, and that would change the answer – avz2611 May 24 '16 at 15:18
• Thanks, I'll modify the proof. – Aravind May 24 '16 at 18:00
• so assuming those three additional inequalities, can you at least prove the requested inequality? If not what progress does this represent? – hkBst May 27 '16 at 10:28 The method to be employed here is exactly the same as in a previous answer:

Let $u = x$ , $v = y^2$ , $w = z^3$ , then $u,v,w \ge 0$ , $u+v+w=1$ and the inequality to be established: $$x^2y+y^2z+z^2x < \frac12 \quad \Longrightarrow \quad f(u,v,w) = u^2v^{1/2}+v\,w^{1/3}+w^{2/3}u < \frac12$$ The maximum of this function inside the equilateral triangle must shown to be less than $1/2$.
There is no symmetry argument, because the latter is effectively destroyed by the "weird" condition $x+y^2+z^3=1$ , as it is called. Another proof without words is attempted by plotting a contour map of the function, as depicted. Levels (nivo) of these isolines are defined (in Delphi Pascal) as:

nivo := min + sqrt(g/grens)*(max-min); { sqrt = square root ; grens = 25 ; g = 0..grens }

The darkness of the isolines is proportional to the (positive) function values; they are almost black near the maximum and almost white near the minimum values. Maximum and minimum values of the function are observed to be:

 4.58251457205350E-0003 < f < 4.85621276951755E-0001 < 1/2

The little $\color{blue}{\mbox{blue}}$ spot is where $\left|f(u,v,w) - \mbox{max}\right|< 0.0002$ . This maximum is close to values found by other people here, but not quite. Perhaps it's interesting to know the position of the maximum as well:

(x,y,z) = ( 5.58304528246164E-0001 , 4.97693736095187E-0001 , 5.78892473099889E-0001 )

I have no idea how to convert these numerical values into something more analytical.

EDIT. What you see is what you get :-) Without doubt. Some decent error analysis reveals that the value of the maximum to be trusted in this answer is : $\;0.48562 \pm 0.00003\;$ , quite in agreement with values given elsewhere (e.g. in the comment by Steve Kass).

I don't think the best bound can be solved for analytically.

using Lagrange multipliers , I tried to maximize $x^2y+y^2z+z^2x$ $\quad$ subject to the constraint $x+y^2+z^3=1$

maximize $g=x^2y+y^2z+z^2x+\lambda(x+y^2+z^3-1)$

$\frac {\partial g}{\partial x}=0:2xy+z^2+\lambda=0$

$\frac {\partial g}{\partial y}=0:x^2+2yz+2\lambda y=0$

$\frac {\partial g}{\partial z}=0:y^2+2zx+3\lambda z^2=0$

$\frac {\partial g}{\partial \lambda}=0:x+y^2+z^3-1=0$

$x\frac {\partial g}{\partial x}=0:2x^2y+z^2x+\lambda x=0$

$y\frac {\partial g}{\partial y}=0:x^2y+2y^2z+2\lambda y^2=0$

$z\frac {\partial g}{\partial z}=0:y^2z+2z^2x+3\lambda z^3=0$

$3(x^2y+y^2z+z^2x)=-\lambda (x+2y^2+3z^3)=(2xy+z^2)(x+2y^2+3z^3)$

If the global maximum occur on $x, y, z \ge 0$ then we can solve $(2xy+z^2)(x+2y^2+3z^3) \ge 0$ for $x, y, z \ge 0$

If no local maximum occur at $x, y,z \ge 0$ then we have to search for farthest points from the point of global minimum.

• It looks like all we have to do is search for the farthest point away from the local minimum that also satisfies the constraint. – ibnAbu May 24 '16 at 7:58

Let $x=0$.

Then $y^2+z^3=1 \Rightarrow z^3=1-y^2$

Seek to maximise $p=y^2z=y^2(1-y^2)^{\frac 13}$

WLOG let $u=y^2$ so that $p=u(1-u)^{\frac 13}$

$\frac {dp}{du}=(1-u)^{\frac 13}-u{\frac 13}(1-u)^{-\frac 23}$

$\frac {dp}{du}=\frac 13(3-4u)(1-u)^{-\frac 23}$

Maximum is at $u=\frac 34$

Thus maximum is $p=\frac 34(\frac 14)^{\frac 13}$

$p \le \frac 34(\frac 14)^{\frac 13}$

$p^3 \le \frac {27}{64}\frac 14$

$p^3 \le \frac {27}{256} < \frac {32}{256}$

$p^3 < \frac {1}{8}$

$p < \frac {1}{2}$

We then have to demonstrate that any increase in $x$ creates corresponding decreases in $y^2$ and $z^2$ such that the other two terms in the expression $x^2y+y^2z+z^2x$ increase more slowly than the decrease in $p$.

• You find a maximum at $p = \frac{3}{4}\left(\frac{1}{4}\right)^{\frac{1}{3}} \approx 0.47247$. However, in the comments to the question, Ewan Delanoy found a point which has value $0.484818$. So I don't think your last paragraph will work? – TastyRomeo May 24 '16 at 12:07
• I was hoping someone else would address that... – tomi May 24 '16 at 12:12
• It also seems odd that the maximum I found is at $\frac {\sqrt 3} 2$. Is there some trig transformation that would make this easier? – tomi May 24 '16 at 12:14
• @stalker2133 look there: math.stackexchange.com/questions/1775498/… – Yuri Negometyanov May 27 '16 at 18:26
• @stalker2133 There are two my answers. Follow link, please. – Yuri Negometyanov May 27 '16 at 19:37

Manipulating the fact that $x^2y+y^2z+z^2x$ is unchanged under renaming of $x-y-z$, we can always assume that $$z \leq y \leq x \ \ \ (*)$$ Why? Because if $x+y^2+z^3=1$ then switch names so that power 3 falls on the smallest, 2 on the middle, and the biggest goes with power 1 makes the sum get smaller

$$x+y^2+z^3 \leq 1 \ \ \ (**)$$

Now under (*) we may be able to find an appropriate bound on

$$x^2y+y^2z+z^2x$$

by replacing some variables with bigger ones. But, unluckily, I have not found what expression will give $1/2$ bound. The crude bound of $x^2y+x^2y+x^2y=3x^2y$ maximized, subject to (**) gives a bound of $\frac{3.16}{25 \sqrt5}$.

• if you switch any two variables, then you do not get the same expression. – hkBst May 27 '16 at 10:26

More idea than answer, but it might be of help.

If we replace $x$ by $x^2$ and $z$ by $z^{2/3}$, the constraint becomes $x^2+y^2+z^2=1$ and the inequality becomes

$$x^4y+y^2z^{2/3}+z^{4/3}x^2\lt{1\over2}$$

This looks worse, but the advantage of constraining $(x,y,z)$ to the unit sphere is that we can switch over to spherical coordinates, writing $x=\sin\theta\cos\phi$, $y=\sin\theta\sin\phi$, and $z=\cos\theta$, and make use of some trig identities. Ultimately, with a final change of variable $u=(\cos\theta)^{2/3}$ and $v=-\cos2\phi$, the inequality to prove (if I've done the algebra correctly) is

$$(1-u^2)^{5/2}\left({(1-v)(1+v)^4\over8}\right)^{1/2}+(u-u^4)(1+v)+(u^2-u^5)(1-v)\lt1\quad\text{for }0\le u,v\le1$$

This "reduces" the problem to the unpleasant chore of finding (or bounding) a function of two variables over the unit square. I'll end by noting that if we focus on the easy portion of the function, $f(u,v)=(u-u^4)(1+v)+(u^2-u^5)(1-v)$, we find that $f$ is maximized when $v=1$ and $u=1/\sqrt4$, and

$$f\left({1\over\sqrt4},1\right)={3\over2\sqrt4}\approx0.94494$$

so the true minimum of the full function is not much less than $1$. (Just a note: I multiplied both sides of the original inequality by $2$ to clear out some denominators. Dividing by $2$ changes $0.94494$ to $0.47247$, which is, as it should be, less than the numerical result $0.48562209920309984177$ reported by Steve Kass in comments below the OP.)