$x,y,z \geqslant 0$, $x+y^2+z^3=1$, 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.

  • 2
    $\begingroup$ 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. $\endgroup$ – Ewan Delanoy May 21 '16 at 18:19
  • 5
    $\begingroup$ Mathematica gives the maximum value of $x^2y+y^2z+z^2x < \frac12$ as constrained to be $0.48562209920309984177$ for $x=0.55811253$, $y=0.49841201$, $z=0.57837131$. $\endgroup$ – Steve Kass May 26 '16 at 3:23
  • $\begingroup$ I think, this method will be interesting too. $\endgroup$ – Yuri Negometyanov Apr 6 '17 at 11:25

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 $$ \genfrac{[}{.}{0}{} {x=0,\quad y=\sqrt{0.75},\quad z=\sqrt[3]{0.25}\approx 0.629991,\quad f\approx 0.47247} {x=0,\quad y=0,\quad z=1,\quad f=0}$$

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 $$x=0.6,\quad y=0,\quad z=\sqrt[3]{0.4}\approx 0.736806,\quad f\approx 0.32573.$$

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: $$ (x,y,z,f)\in\left[\genfrac{}{}{0pt}{0} {(0.5581125,\ 0.4984120,\ 0.5783713,\ 0.4856221)} {(0.1636702,\ 0.7619816,\ 0.6347238,\ 0.4548812)} \right. $$

  • $\begingroup$ Less time consuming way to set out as a separate answer. $\endgroup$ – Yuri Negometyanov May 27 '16 at 12:43
  • $\begingroup$ The problem with this is that the equations were solved using some solver (approximations) and so cant represent exact solution $\endgroup$ – ibnAbu May 27 '16 at 18:49
  • $\begingroup$ Worst there is no proof that the approximate solutions give global maximum. $\endgroup$ – ibnAbu May 27 '16 at 18:50
  • $\begingroup$ @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.))) $\endgroup$ – Yuri Negometyanov May 27 '16 at 19:44
  • $\begingroup$ I checked the link but I wonder why we are using determinant to tell about solutions of non- linear equations. $\endgroup$ – 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 $\dfrac{A+B+C}{4}$.

  • 3
    $\begingroup$ Isn't a+b+c=2, and that would change the answer $\endgroup$ – avz2611 May 24 '16 at 15:18
  • $\begingroup$ Thanks, I'll modify the proof. $\endgroup$ – Aravind May 24 '16 at 18:00
  • 1
    $\begingroup$ so assuming those three additional inequalities, can you at least prove the requested inequality? If not what progress does this represent? $\endgroup$ – hkBst May 27 '16 at 10:28

enter image description here

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.

  • $\begingroup$ It looks like all we have to do is search for the farthest point away from the local minimum that also satisfies the constraint. $\endgroup$ – 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 $p$.

  • 3
    $\begingroup$ 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? $\endgroup$ – TastyRomeo May 24 '16 at 12:07
  • 1
    $\begingroup$ I was hoping someone else would address that... $\endgroup$ – tomi May 24 '16 at 12:12
  • $\begingroup$ 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? $\endgroup$ – tomi May 24 '16 at 12:14
  • $\begingroup$ @stalker2133 look there: math.stackexchange.com/questions/1775498/… $\endgroup$ – Yuri Negometyanov May 27 '16 at 18:26
  • $\begingroup$ @stalker2133 There are two my answers. Follow link, please. $\endgroup$ – 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 $\frac{3.16}{25 \sqrt5}$.

  • $\begingroup$ if you switch any two variables, then you do not get the same expression. $\endgroup$ – 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


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 $u=1/\sqrt[3]4$, and


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.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.