Find the maximum of the value $F=x^3y+y^3z+z^3x$

Question

let $x,y,z$ be real number.if $x+y+z=3$,show that $$x^3y+y^3z+z^3x\le \dfrac{9(63+5\sqrt{105})}{32}$$

and the inequality $=$,then $x=?,y=?,z=?$

I can solve if add $x,y,z\ge 0$,also see: Calculate the maximum value of $x^3y + y^3z + z^3x$ where $x + y + z = 4$ and $x, y, z \ge 0$.

But for real $x,y,z$ I can't solve it

Answer 1

We have \begin{align*} &x^3y + y^3z + z^3 x\\ =\,& \frac{(x + y + z)(x^3 + y^3 + z^3) - (x^4 + y^4 + z^4)}{2} - \frac{x + y + z}{2}(x - y)(y - z)(z - x)\\ =\, & \frac{3(x^3 + y^3 + z^3) - (x^4 + y^4 + z^4)}{2} - \frac{3}{2}(x - y)(y - z)(z - x)\\ \le\, & \frac{3(x^3 + y^3 + z^3) - (x^4 + y^4 + z^4)}{2} + \frac{3}{2}\sqrt{(x - y)^2(y - z)^2(z - x)^2}. \tag{1} \end{align*} It suffices to prove that $$\frac{9(63+5\sqrt{105})}{32} - \frac{3(x^3 + y^3 + z^3) - (x^4 + y^4 + z^4)}{2} \ge \frac{3}{2}\sqrt{(x - y)^2(y - z)^2(z - x)^2}.$$ Note that \begin{align*} &3(x^3 + y^3 + z^3) - (x^4 + y^4 + z^4)\\ =\,& -\frac{1}{64}(8x^2 - 12x - 9)^2 -\frac{1}{64}(8y^2 - 12y - 9)^2 -\frac{1}{64}(8z^2 - 12z - 9)^2\\ &\quad + \frac{27}{8}(x + y + z) + \frac{243}{64}\\ \le\, & \frac{27}{8}\cdot 3 + \frac{243}{64}\\ <\, & 2\cdot \frac{9(63+5\sqrt{105})}{32}. \end{align*} It suffices to prove that \begin{align*} & \left[\frac{9(63+5\sqrt{105})}{32} - \frac{3(x^3 + y^3 + z^3) - (x^4 + y^4 + z^4)}{2}\right]^2\\ \ge\, & \frac{9}{4}(x - y)^2(y - z)^2(z - x)^2. \tag{2} \end{align*}

Let $p = x + y + z = 3$, $q = xy + yz + zx$ and $r = xyz$.

Denote $Q = \frac{9(63+5\sqrt{105})}{32}$. (2) is written as $$\left(Q - \frac92\,q + \frac32\,r + {q}^{2}\right)^2 \ge \frac{9}{4}(-4\,{q}^{3}+9\,{q}^{2}+54\,qr-27\,{r}^{2}-108\,r)$$ or $$63\,{r}^{2} + 3\, \left(Q + {q}^{2} - 45\,q + 81 \right) r + {Q}^{2}+ \left( 2 \,{q}^{2}-9\,q \right) Q+{q}^{4} \ge 0 $$ or \begin{align*} &\frac{1}{28}\, \left( {q}^{2}+Q-45\,q+42\,r+81 \right) ^{2} + \,{\frac { 384}{117\,\sqrt {105}+721}}\left( 8\,q+9 +3\,\sqrt {105} \right) ^{2}\\ &\quad + {\frac {3 }{114688}}\, \left( 24\,q+13-9\,\sqrt {105} \right) ^{2} \left( 8\,q+9+3\,\sqrt {105} \right) ^{2} \ge 0 \end{align*} which is clearly true.

We are done.

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Remarks:

Actually, $\frac{9(63+5\sqrt{105})}{32}$ is the maximum.

Let $x_0 > y_0 > z_0$ be the three real roots of the cubic $${u}^{3}-3\,{u}^{2}+ \left( -{\frac {9}{8}}-\frac38\,\sqrt {105} \right) u+ {\frac {63}{16}}+{\frac {51}{112}}\,\sqrt {105} = 0.$$

Let $p_0 = x_0 + y_0 + z_0, q_0 = x_0y_0 + y_0z_0 + z_0x_0, r_0 = x_0y_0z_0$. Then $$p_0 = 3, ~ q_0 = -{\frac {9}{8}}-\frac38\,\sqrt {105}, ~ r_0 = - {\frac {63}{16}} - {\frac {51}{112}}\,\sqrt {105}.$$

We have (see (1)) \begin{align*} &x_0^3y_0 + y_0^3z_0 + z_0^3 x_0\\ =\, & \frac{3(x_0^3 + y_0^3 + z_0^3) - (x_0^4 + y_0^4 + z_0^4)}{2} + \frac{3}{2}\sqrt{(x_0 - y_0)^2(y_0 - z_0)^2(z_0 - x_0)^2}\\ =\, & \frac{9}{2}q_0 - \frac{3}{2}r_0 - q_0^2 + \frac{3}{2}\sqrt{-4\,q_0^{3} + 9\,q_0^{2} + 54\,q_0 r_0 - 27\, r_0^{2}-108\,r_0}\\ =\, & \frac{9(63+5\sqrt{105})}{32}. \end{align*}

Answer 2

We can use the $uvw$'s technique here.

Indeed, let $\dfrac{9(63+5\sqrt{105})}{32}=k$, $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v^2$ can be negative, and $xyz=w^3$.

Thus, we need to prove that: $$ku^4\geq\sum_{cyc}x^3y$$ or $$2ku^4-\sum_{cyc}(x^3y+x^3z)\geq\sum_{cyc}(x^3y-x^3z)$$ or $$2ku^4-(27u^2v^2-18v^4-3uw^3)\geq3u(x-y)(x-z)(y-z).$$ We'll prove that $2ku^4-(27u^2v^2-18v^4-3uw^3)\geq0.$

Indeed, this inequality is a linear inequality of $w^3$, which by $uvw$ says that it's enough to prove this inequality for equality case of two variables.

Let $y=z=1$.

We need to prove that $$\frac{2k}{81}(x+2)^4-2x^3-2x-2\geq0$$ and since $2k>27$, it's enough to prove that $$(x+2)^4\geq6(x^3+x+1)$$ or $$x^4+2x^3+24x^2+26x+10\geq0$$ or $$x^2(x+1)^2+23x^2+26x+10\geq0,$$ which is obvious.

Id est, it's enough to prove that: $$(2ku^4-(27u^2v^2-18v^4-3uw^3))^2\geq9u^2\prod_{cyc}(a-b)^2$$ or $$(2ku^4-(27u^2v^2-18v^4-3uw^3))^2\geq243u^2(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6),$$ which is a quadratic inequality of $w^3$.

Now, prove that $\Delta\leq0.$

Can you end it now?

Answer 3

It is not surprizing that Mathematica does not produce explicit expressions for $(x,y,z)$.

Using it with the restriction $x>y>z$, it gives that $y$ and $z$ are respectively the fourth and the first roots of the equation $$56 t^6-336 t^5+378 t^4+819 t^3-2079 t^2+1512 t-351=0$$

Using @River Li's comment, $(x,y,z)$ are the solutions of the cubic $$u^3-3u^2-\frac{3}{8} \left(3+\sqrt{105}\right)u+\frac{3}{112} \left(147+17 \sqrt{105}\right)=0$$ and, using the trigonometric method, they are $$x=1+\sqrt{\frac{11+\sqrt{105}}{2} } \cos \left(\frac{1}{3} \cos ^{-1}\left(-\frac{91+9 \sqrt{105}}{7 \sqrt{2} \left(11+\sqrt{105}\right)^{3/2}}\right)\right)=3.71382\cdots$$ $$y=1-\sqrt{\frac{11+\sqrt{105}}{2} } \sin \left(\frac{\pi }{6}-\frac{1}{3} \cos ^{-1}\left(-\frac{91+9 \sqrt{105}}{7 \sqrt{2} \left(11+\sqrt{105}\right)^{3/2}}\right)\right)=1.20642\cdots$$ $$z=1-\sqrt{\frac{11+\sqrt{105}}{2} } \sin \left(\frac{\pi }{6}+\frac{1}{3} \cos ^{-1}\left(-\frac{91+9 \sqrt{105}}{7 \sqrt{2} \left(11+\sqrt{105}\right)^{3/2}}\right)\right)=-1.92024\cdots$$

Trying to convert to radicals would be more than problematic.

Edit

After a series of mistakes, using @River Li's comments, the factorization of the sextic as the product of two cubic polynomials $$(t^3 + b_2t^2 + b_1t + b_0)\,(t^3 + a_2t^2 + a_1t + a_0)$$gives $$b_2=-3 \qquad \qquad b_1=-\frac{3}{8} \left(3-\sqrt{105}\right)\qquad \qquad b_0=\frac{3}{112} \left(147-17 \sqrt{105}\right)$$ $$a_2=-3 \qquad \qquad a_1=-\frac{3}{8} \left(3+\sqrt{105}\right)\qquad \qquad a_0=\frac{3}{112} \left(147+17 \sqrt{105}\right)$$

Answer 4

A SOS (Sum of Squares) solution with computer:

Denote $K = \frac{9(63+5\sqrt{105})}{32}$.

We have $$ K\left(\frac{x + y + z}{3}\right)^4 - (x^3y + y^3z + z^3x) = \alpha_1 f_1^2 + \alpha_2 f_2^2 + \alpha_3 f_3^2 $$ where \begin{align*} \alpha_1 &= \frac{567 - 16K}{13778100} > 0, \\[5pt] \alpha_2 &= \frac{472K - 5607}{292287082500} > 0,\\[5pt] \alpha_3 &= \frac{998K - 29538}{1491260625} > 0, \\[8pt] f_1 &= \left( 10\,{x}^{2}+20\,xy+20\,xz-14\,{y}^{2}+44\,yz-14\,{z}^{2} \right) K \\ &\qquad -405\,xy+459\,{y}^{2}-999\,yz+459\,{z}^{2},\\[8pt] f_2 &= \left( 944\,xy-112\,xz-592\,{y}^{2}+1440\,yz-480\,{z}^{2} \right) K \\ &\qquad - 22239\,xy+8442\,xz+21987\,{y}^{2}-40635\,yz+17325\,{z}^{2},\\[8pt] f_3 &= \left( 16\,{y}^{2}-48\,yz \right) K+45\,xz-261\,{y}^{2}+918\,yz-405\, {z}^{2}. \end{align*}

Answer 5

Here is a solution given with computer aid, here sage. (The OP also uses a computational approach.) The plan is as follows. We use the parametrization $x=1+a$, $y=1+b$, $z=1-a-b$ to get rid of the sum condition, the function $E$ (notation) to be maximized is now a polynomial function of degree four defined for all $(a,b)\in\Bbb R^2$. It is explicitly: $$ \begin{aligned} E &= E_4+E_3+E_2+E_1+E_0\ ,\\[2mm] E_4 &= -(\ a^{4} + 2 a^{3} b + 3 a^{2} b^{2} + 2 a b^{3} + b^{4}\ ) =-(\ a^2 + ab + b^2\ )^2\le 0\ ,\\ E_3 &= 3 (\ a^{3} + 2 a^{2} b - a b^{2} - b^{3}\ )\ ,\\ E_2 &= 3(\ a^2 + ab + 3b^2\ )\ ,\\ E_1 &= 0\ ,\\ E_0 &= 3\ . \end{aligned} $$ It is clear that $E/E_4$ converges to $-1$ for $(a,b)$ converging in norm to infinity, so the given function $E$ has an absolute maximum, that can be found by finding the critical points of $E$, i.e. solutions $(a_0, b_0)$ of the algebraic system with two equations $$ \tag{$*$} E'_a=E'_b=0\ , $$ plugging them in $E$ one by one, and taking the maximal value among them. It is easy to solve this system using computing power. The values are explicit (exact) algebraic numbers, we know for instance the minimal polynomial over $\Bbb Q$ of $a_0$, $b_0$, and it turns out, it is the same minimal polynomial. We also know a numerical approximation for the possible values of the optimal point $(a_0,b_0)$, and this is enough to "put the hands" on the point. The engine also computes $E(a_0,b_0)$ for all critical points, for each such value we have again the minimal polynomial and the approximation, so we "know" the point. By chance, the optimal value $\max E=E(a_0,b_0)$ of $E$ in an optimal point is the solution of a polynomial of degree two, so this value can be easily written down. Not so simple is the explicit formula for the optimal point, we need to solve an equation of degree three, but this can also be done. This was the plan, let us realize it step by step.

To have a picture in this introductory part, here is a plot of $\max(E, 25)$ for the variable $(a,b)$ running in $[-4,4]^2$. (The bottom truncation with $25$ was needed, else there is a steep fall caused by the $E_4$ term in $E$.)

$$ %sage: plot3d( lambda a, b: E(1+a, 1+b, 1-a-b), (-4, 4), (-4, 4), %plot_points=300 ) %Launched html viewer for Graphics3d Object %sage: def E(x, y, z): %....: value = x^3*y + y^3*z + z^3*x %....: return value if value > 25 else 25. %....: %sage: var('a,b'); %sage: def E(x, y, z): %....: value = x^3*y + y^3*z + z^3*x %....: return value if value > 25 else 25. %....: %sage: $$

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We start by passing from the expression $x^3y+y^3z+z^3x$, which should be maximized with the given constraint, to a polynomial expression $E$, obtained by using the mentioned parametrization in terms of $(a,b)\in\Bbb R^2$. Recall that $\sup E$ is not taken "at infinity", so $\sup E=\max E$ is a global maximum, which can be computed by considering all critical points of $E$, taking the values in these points, taking the maximum.

Solving $(*)$ is done below by associating the ideal generated by $E'_a, E'_b$ in $\Bbb Q[a,b]$, then asking for its "variety". From the human perspective, it is also interesting to compute the result of eliminating one variable, here $b$, then $a$, from the two equations $(*)$. Time for the code:

R.<a, b> = PolynomialRing(QQ)
x, y, z = 1 + a, 1 + b, 1 - a -b

E = x^3*y + y^3*z + z^3*x
J = R.ideal([diff(E, a), diff(E, b)])
Ja = J.elimination_ideal(b)
Jb = J.elimination_ideal(a)

print(f'E is the polynomial:\n{E}\n')
print(f'The ideal J is:\n{J}\n')
print(f'Eliminating b we obtain the ideal:\n{Ja}\n')
print(f'Eliminating a we obtain the ideal:\n{Jb}\n')

And we obtain:

E is the polynomial:
-a^4 - 2*a^3*b - 3*a^2*b^2 - 2*a*b^3 - b^4
    + 3*a^3 + 6*a^2*b - 3*a*b^2 - 3*b^3
    + 3*a^2 + 3*a*b + 3*b^2
    + 3

The ideal J is:
Ideal (-4*a^3 - 6*a^2*b - 6*a*b^2 - 2*b^3 + 9*a^2 + 12*a*b - 3*b^2 + 6*a + 3*b, 
       -2*a^3 - 6*a^2*b - 6*a*b^2 - 4*b^3 + 6*a^2 - 6*a*b - 9*b^2 + 3*a + 6*b) 
    of Multivariate Polynomial Ring in a, b over Rational Field

Eliminating b we obtain the ideal:
Ideal (56*a^7 - 462*a^5 + 91*a^4 + 126*a^3 - 21*a^2 - a)
    of Multivariate Polynomial Ring in a, b over Rational Field

Eliminating a we obtain the ideal:
Ideal (56*b^7 - 462*b^5 + 91*b^4 + 126*b^3 - 21*b^2 - b)
    of Multivariate Polynomial Ring in a, b over Rational Field

Results were manually rearranged. We have a more or less direct access to the solutions of $(*)$ by asking for J.variety(), however, we also need to have the roots of the above generator of Ja in the ground field. (Above, it was $\Bbb Q$, we must extend.) We extend, so that the base field becomes $\bar{\Bbb Q}=$QQbar, since it "knows the roots". Then we ask for the values of $E$ in the critical points. These values are algebraic known numbers, there is no numerical step involved, to have them "exactly", we also ask for the minimal polynomial.

V = J.variety(ring=QQbar)
print('E-values in rational points of the variety defined by J / QQbar:')
for dic in V:
    a0, b0 = dic[a], dic[b]
    print(f'E( {a0}, {b0} ) = {E(a0, b0)} with minpoly = {E(a0, b0).minpoly()}')

Results:

E-values in rational points of the variety defined by J / QQbar:
E( 0, 0 ) = 3 with minpoly = x - 3
E( 0.2064248138599934?, -2.920240652156161? ) = 32.12852451463069? with minpoly = x^2 - 567/16*x + 1701/16
E( 0.549771246579960?, -0.5109092345889097? ) = 3.308975485369315? with minpoly = x^2 - 567/16*x + 1701/16
E( -0.5109092345889097?, -0.03886201199105012? ) = 3.308975485369315? with minpoly = x^2 - 567/16*x + 1701/16
E( 2.713815838296168?, 0.2064248138599934? ) = 32.12852451463069? with minpoly = x^2 - 567/16*x + 1701/16
E( -0.03886201199105012?, 0.549771246579960? ) = 3.308975485369315? with minpoly = x^2 - 567/16*x + 1701/16
E( -2.920240652156161?, 2.713815838296168? ) = 32.12852451463069? with minpoly = x^2 - 567/16*x + 1701/16

So the maximal value $\max E$ of $E$ on $\Bbb R^2$ is the root of the polynomial $16x^2 - 567x + 1701$ near $32$, which is: $$ \color{blue}{ \max E=\frac 1{32}(567+45\sqrt{105})=\frac 9{32}(63+5\sqrt{105})\ . } $$ This is already a computer aided proof.

$\square$

(Sage makes exact computations inside QQbar.) We have the maximum. The OP also asks for the point(s) $(x,y,z)$ realizing this maximal value.

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We note that for an optimal point $(a_0, b_0)$, the point $(b_0,c_0)$ with $c_0=-(a_0+b_0)$ is also optimal. This is so, since if $(x_0,y_0,z_0)=(1+a_0, 1+b_0, 1-a_0-b_0)$ is optimal for $x^3x+y^3z+z^3x$, if( and only i)f $(y_0,z_0,x_0)=(1+b_0, 1+c_0, 1-b_0-c_0)$ is optimal.

This explains the structure of the extremal points:

V = J.variety(ring=QQbar)
print('Optimal E-values in rational points of the variety defined by J / QQbar:')
for dic in V:
    a0, b0, c0 = dic[a], dic[b], -(dic[a] + dic[b])
    if E(a0, b0) > 32:
        print(f'a0 = {a0} : b0 = {b0} : c0 = -(a0 + b0) = {c0}')

... which gives:

Optimal E-values in rational points of the variety defined by J / QQbar:
a0 = 0.2064248138599934? : b0 = -2.920240652156161? : c0 = -(a0 + b0) = 2.713815838296168?
a0 = 2.713815838296168? : b0 = 0.2064248138599934? : c0 = -(a0 + b0) = -2.920240652156161?
a0 = -2.920240652156161? : b0 = 2.713815838296168? : c0 = -(a0 + b0) = 0.2064248138599934?

Which is the equation satisfied by these points. We compute starting from the values a0, b0, c0 from the last loop:

sage: a0, b0, c0
(0.2064248138599934?, -2.920240652156161?, 2.713815838296168?)
sage: a0 + b0 + c0
0
sage: a0*b0 + b0*c0 + c0*a0
-7.96760653723485?
sage: (a0*b0 + b0*c0 + c0*a0).minpoly()
x^2 + 33/4*x + 9/4
sage: a0*b0*c0
-1.635915686550325?
sage: (a0*b0*c0).minpoly()
x^2 + 13/8*x - 1/56

The two numbers above are $p = a_0b_0+b_0c_0+c_0a_0=-\frac 38(11+\sqrt{105})$, $q=-\frac 1{112}(91+9\sqrt{105})$. One can obtain this polynomial $x^3+px-q$ also as a factor of the generator of Ja, e.g.

K.<w> = QuadraticField(105)    # w is sqrt(105)
R.<a> = PolynomialRing(K)
factor(56*a^7 - 462*a^5 + 91*a^4 + 126*a^3 - 21*a^2 - a)

And we get:

(56) * a * (a^3 + (-3/8*w - 33/8)*a + 9/112*w + 13/16)
         * (a^3 + ( 3/8*w - 33/8)*a - 9/112*w + 13/16)

(manually adjusted).

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Let us get the explicit formula for the roots of the equation $X^3+pX-q=0$. We try to write $p,q$ in the form $p=-3ST$, $-q=-S^3-T^3$, so that $$ X^3+pX-q= X^3 -3ST\; X - S^3-T^3 =(X-S-T)(\dots)\ , $$ and the roots are easily extracted now. From the above, we have the sum $q$ and the product $-(p/3)^3$ of $S^3$ and $T^3$, so these values are... $$ \frac 12\left(-q\pm\sqrt{q^2+\frac 4{27}p^3}\right) \ . $$

This leads to the explicit (almost) radical formula for the roots: $$ \color{blue}{a_0,b_0,c_0= -2\Re\left(\ \frac 1{224} \left( 91 + 9\sqrt{105} + 3i\sqrt{42(1199 + 117\sqrt{105})} \right) \ \right)^{1/3} \ .} $$ Here, there are three algebraic roots to be built over $\Bbb C$. Let us check numerically the formula:

w = sqrt(105)
A0 = -(                2 * ( 1/224 * (91 + 9*w + 3*i*sqrt(42*(1199 + 117*w))) )^(1/3) ).real()
B0 = -( (-1 + i*sqrt(3)) * ( 1/224 * (91 + 9*w + 3*i*sqrt(42*(1199 + 117*w))) )^(1/3) ).real() 
C0 = -( (-1 - i*sqrt(3)) * ( 1/224 * (91 + 9*w + 3*i*sqrt(42*(1199 + 117*w))) )^(1/3) ).real()

print(f'A0 = {A0.n()}')
print(f'B0 = {B0.n()}')
print(f'C0 = {C0.n()}')

And we get (up to reordering) a numerical match with the already computed exact algebraic numbers a0, b0, c0:

A0 = -2.92024065215616
B0 = 2.71381583829617
C0 = 0.206424813859994

One has to add $1$ to get the corresponding values for $x,y,z$.

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Note: One can use Lagrange multiplicators to solve the problem directly using $x,y,z$ (and the multiplier parameter $t$). Computer solution for this approach:

R.<x,y,z,t> = PolynomialRing(QQ)
f = x^3*y + y^3*z + z^3*x
h = (x + y + z - 3)
F = f + t*h
J = R.ideal([diff(F, v) for v in (x, y, z, t)])
V = J.variety(ring=QQbar)

for point in V:
    x0, y0, z0 = point[x], point[y], point[z]
    value = x0^3*y0 + y0^3*z0 + z0^3*x0
    if value > 20:
        print( f'x0 = {x0}')
        print( f'y0 = {y0}')
        print( f'z0 = {z0}')
        print( f'value = {value}\n')

And we get:

x0 = 3.713815838296168?
y0 = 1.206424813859994?
z0 = -1.920240652156161?
value = 32.12852451463069?

x0 = -1.920240652156161?
y0 = 3.713815838296168?
z0 = 1.206424813859994?
value = 32.12852451463069?

x0 = 1.206424813859994?
y0 = -1.920240652156161?
z0 = 3.713815838296168?
value = 32.12852451463069?

This would be a second (computer assisted) solution.

$\square$

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Note that (cyclic and) only cyclic permutations of a solution is again a solution. So restricting to either $x<y<z$ or $y <x<z$ is a priori not allowed, both cases have to be considered.