Maximization with constraints

Let $x,y,z$ be the real roots of a monic, cubic polynomial.

Then, $P(t) = (t-x)(t-y)(t-z) = t^3+At^2+Bt+C$ .

By expanding the left side, and equating coefficients, we get:

$A = -(x+y+z) = 0$ ,

$B = xy+yz+zx = \dfrac{1}{2}\left((x+y+z)^2-(x^2+y^2+z^2)\right) = -\tfrac{1}{2}$ ,

$C = -xyz$ .

We want the maximum value of $x^2y^2z^2 = C^2$ such that the cubic equation $P(t) = t^3 - \dfrac{1}{2}t+C = 0$ has three real roots.

Since $P'(t) = 3t^2 - \tfrac{1}{2} = 0 \leadsto t = \pm \tfrac{1}{\sqrt{6}}$ , and $P(t)$ is a cubic with positive leading coefficient, we know that $P(t)$ has a local minimum/maximum at $\left(\pm\tfrac{1}{\sqrt{6}}, P\left(\pm\tfrac{1}{\sqrt{6}}\right)\right) = \left(\pm\tfrac{1}{\sqrt{6}},\mp\tfrac{1}{3\sqrt{6}}+C\right)$ .

Note that a cubic polynomial has three real roots iff the local maximum value is non-negative and the local minimum value is non-positive. Therefore, we need $\tfrac{1}{3\sqrt{6}} + C \ge 0$ and $-\tfrac{1}{3\sqrt{6}} + C \le 0$ .

So, $-\tfrac{1}{3\sqrt{6}} \le C \le \tfrac{1}{3\sqrt{6}}$ , and thus, $x^2y^2z^2 = C^2 \le \tfrac{1}{54}$ . Equality is achievable for $(x,y,z) = \left(\mp\tfrac{2}{\sqrt{6}}, \pm\tfrac{1}{\sqrt{6}}, \pm\tfrac{1}{\sqrt{6}}\right)$ and permutations.

Thus, the maximum value of $C^2 = x^2y^2z^2$ is $\tfrac{1}{54} = \tfrac{a}{b} \leadsto a+b = \boxed{55}$ .

$x + y + z =0$ $\Rightarrow z^2 = (x + y)^2$

$\Rightarrow x^2 + y^2 + z^2 = 2x^2 + 2y^2 + 2xy= 1$

$\Rightarrow x^2 + y^2 + xy = \frac{1}{2}$ and $(x+y)^2 = z^2 = \frac{1}{2} + xy$

Using $x^2 + y^2 \geq 2xy$ in above equation

$\Rightarrow xy \leq \frac{1}{6}$

Now , $x^2y^2z^2 = x^2y^2(x+y)^2 = (x^2y^2)(\frac{1}{2} + xy)$

Since , $xy \leq \frac{1}{6}$

$\Rightarrow x^2y^2z^2 \leq \frac{1}{54}$ . $\Rightarrow a + b = \fbox{55}$

We claim that the maximum is $\frac{1}{54}$ . This could be achieved when $x=y=\frac{\sqrt{6}}{6}, z=\frac{-\sqrt{6}}{3}$ , any permutation or the result if you multiply each of $x, y, z$ by $-1$ .

We would disregard the case where any of $x, y, z$ is $0$ , because in that case, $x^2y^2z^2$ will be $0$ , but we have shown that $\frac{1}{54}$ , which is greater than $0$ , is achievable.

By the pigeonhole principle, at least two of $x, y, z$ are either both negative or nonnegative. Since we are disregarding the case where any of them is $0$ , without loss of generality, we assume that $x, y$ are positive. Then $z=-x-y$ . Note that we are now maximizing $x^2y^2z^2=x^2y^2(-x-y)^2$ .

Plugging $z=-x-y$ in $x^2+y^2+z^2=1$ gives $x^2+y^2+(-x-y)^2=1$ , so $2x^2+2y^2+2xy=1$ , so $x^2+y^2+xy=\frac{1}{2}$ . By the AM-GM inequality, $x^2+y^2 \ge 2xy$ . Hence $xy \le (\frac{1}{2})(\frac{1}{3})=\frac{1}{6}$ . Thus $(x+y)^2=x^2+2xy+y^2 \le \frac{1}{2}+\frac{1}{6}=\frac{2}{3}$ , so $x+y \le \sqrt{\frac{2}{3}}=\frac{\sqrt{6}}{3}$ .

Hence $xy(x+y) \le (\frac{1}{6})(\frac{\sqrt{6}}{3})=\frac{\sqrt{6}}{18}$ . Thus $x^2y^2z^2=x^2y^2(x+y)^2=(x^2y^2(-x-y))^2 \le (\frac{\sqrt{6}}{18})^2=\frac{6}{324}=\frac{1}{54}$ .

Since the largest possible value of $x^2y^2z^2$ is $\frac{1}{54}$ , $a=1$ and $b=54$ , so $a+b=1+54=\boxed{55}$ .

Notice $z=-(x+y)$ . Via substitution, $x$ and $y$ must satisfy $x^2 + y^2 + (x+y)^2 = 1$ . By symmetry, the maximum of $x^2y^2z^2 = x^2y^2(x+y)^2$ occurs when $x=y$ . Thus $x = y = \dfrac{1}{\sqrt{6}}$ , hence $z=\dfrac{2}{\sqrt{6}}$ . The value of $x^2y^2z^2$ is therefore $\dfrac{4}{216} = \boxed{\dfrac{1}{54}}$

This is quite inelegant and involves calculus, but I could not seem to make the geometry angle work for me.

Treat $x,y,z$ as the roots of a cubic equation with leading coefficient 1. The given information implies that the sum of the roots is 1.

We also have $(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$ . So we must have $xy+yz+zx=-1/2$ We also write $c=xyz$

Putting these bits of information together, we know the equation must be of the form $u^3-\frac{1}{2}u-c=0$

Rearranging the equation we have $u^3=\frac{1}{2}u+c$ . So we are looking for intersections between a line of slope $\frac{1}{2}$ and the graph of $u^3$ . The question says that x,y,z are distinct real numbers, so we need the cubic to have 3 real roots (not necessarily distinct).

Clearly changing $c$ only changes the intercept of the line, and we want the intercept to be as large as possible. Sketching the graphs, it is clear to see that this is the case when the line with slope $\frac{1}{2}$ is tangent to the cubic.

Differentiating and setting the slope to $\frac{1}{2}$ we get $3u^2=\frac{1}{2}$ and $u=\pm \frac{1}{\sqrt{6}}$ .

At this value of $u$ both the line and the curve have the same y coordinate, so $-\frac{1}{6\sqrt{6}}=\frac{1}{2}\times\frac{1}{6\sqrt{6}}+c$

$\therefore c=\frac{1}{3\sqrt{6}}$ and $c^2=\frac{1}{54}$

We already know $x+y+z=0$ and $xy+yz+zx=-1/2$ . Now let $u=xyz$ and consider the cubic equation

$t^3-\frac{1}{2}t-u=0$

Recall that for the cubic equation $t^3+pt+q=0$ we have the discriminant

$\Delta=\frac{q^2}{4}+\frac{p^3}{27}$

, where the equation has real roots iff $\Delta\le 0$ . So for our equation, we have

$\Delta=-\frac{1}{216}+\frac{u^2}{4}\le0\Longrightarrow x^2y^2z^2=u^2\le \frac{1}{54}$

Take x as rcos(t) and y as rsin(t) where r=sqrt(1-z^2) . This is got from the second equation . Now substitute this in the first equation where we get sin(t) + cos(t) as a function of z. Now square the equation and get sin(t)xcos(t) . Now x and y as a function of z only . Now substitute into the third equation which becomes a function of z only . Now differentiate once to get the corresponding value of z ( z^2 is 1/6) and substitute this into the third equation to get maximum value .

Since x+y+z=0 either x=y=z=0, 2 of the variables are positive and 1 negative, or vice versa. The former case doesn't hold for the second condition, and the second and third cases are synonymous. WLOG let x+y=z.

Now plugging that into the second condition and simplifying we obtain x^2+y^2+xy=1/2. We want to maximize x^2y^2(x+y)^2.

Now, consider the quadratic a^2+ba+c with real roots x and y. The sum of the roots (x+y) is -b, and the product (xy) is c. Therefore we can rewrite as having b^2-c=1/2 and wanting to maximize b^2c^2.

From AM-GM, note that $\frac{b^2-c/2-c/2}{3} \ge \sqrt[3]{\frac{b^2c^2}{4}}$ . This boils down to $(\frac{1/2}{3})^3 \ge b^2c^2/4$ so the maximum value of b^2c^2 is 1/54 and the answer is 55.

From the given conditions, we have $z=-(x+y)$ so that $x^2+y^2 + (x+y)^2 = 1.$ Expanding and collecting terms, we have $x^2+y^2 + xy = \frac{1}{2}.$ By AM-GM inequality, we obtain $x^2+y^2 \ge 2xy$ with equality if and only if $x=y.$ From the previous equation, we then have $2xy + xy \le \frac{1}{2}$ or $xy \le \frac{1}{6}.$ Since $x^2+y^2 = (x+y)^2-2xy,$ from the previous equation, we also have $(x+y)^2-xy = \frac{1}{2}$ or $(x+y)^2 = \frac{1}{2}+xy.$ This implies $\frac{1}{2}+xy \ge 0$ or $xy \ge -\frac{1}{2}$ and $x^2y^2z^2 = (xy)^2(x+y)^2 = (xy)^2(\frac{1}{2}+xy).$ Let $t = xy$ . We want to maximize the function $f(t) = t^2(\frac{1}{2}+t), \quad -\frac{1}{2} \le t\le \frac{1}{6}.$ With a littile help from calculus, we find $f'(t) = t(3t+1)$ so that relative extreme values occur at $t=0$ or $t=-\frac{1}{3}.$ Since $f$ is continuous on $[-\frac{1}{2},\frac{1}{6}],$ the absolute extreme values have to occur at one of the following points: $-\frac{1}{2}, -\frac{1}{3},0,\frac{1}{6}.$ We compare the values of $f(t)$ at these points and find that $f(-\frac{1}{3}) = f(\frac{1}{6}) = \frac{1}{54}$ is the maximum value. Hence the maximum of $x^2y^2z^2$ is $\boxed{\frac{1}{54}},$ attained when, for instance, $x=y = \frac{1}{\sqrt{6}}$ and $z=-\frac{2}{\sqrt{6}}.$

To simplify the problem, we will eliminate $x$ by writing $x$ as $(-y-z)$ . The problem now becomes: given that $(-y-z)^2+y^2+z^2=1$ , find the maximum value of $(-y-z)^2y^2z^2$

The constraint is equivalent to $2y^2+2z^2+2yz=1$ , or $y^2+z^2+yz=\frac{1}{2}$ .

We have to find the maximum value of $(-y-z)^2y^2z^2=(y^2+2yz+2z^2)(yz)^2$ . From the constraint we have $y^2+z^2+yz=\frac{1}{2}$ , so $y^2+z^2+2yz=\frac{1}{2}+yz$ . Therefore we need to maximize $(\frac{1}{2}+yz)(yz)^2$ .

Now, since $y^2+z^2+yz=\frac{1}{2}$ , and we know that $y^2+z^2 \geq 2yz$ , $\frac{1}{2} \geq 2yz +yz =3yz$ . Thus, $yz\leq \frac{1}{6}$

Hence the maximum value we need to find is $(\frac{1}{2}+yz)(yz)^2 \leq \left(\frac{1}{2} +\frac{1}{6}\right) \left(\frac{1}{6}\right)^2 =\frac{1}{54}$ . The equality sign occurs when $y=z=\frac{1}{\sqrt{6}},x=\frac{-2}{\sqrt{6}}$ .

Therefore, $a+b=1+54=\fbox{55}$

We have: $2yz=(y+z)^2-(y^2+z^2)$ So $2yz=2x^2-1$ Rewrite $x^2y^2z^2=x^6-x^4+\frac{1}{4}x^2$ . Applying : $(y+z)^2\geq 4yz$ We get: $0\leq x^2\leq \frac{2}{3}$ . Let $x^2=t$ ;We see $f(t)=t^3-t^2+\frac{1}{4}t$ in the interval $0\leq t\leq \frac{2}{3}$ . $f(0)=f(\frac{1}{2})=0;f(\frac{1}{6})=f(\frac{2}{3})=\frac{1}{54}$ So max of $f(t)=\frac{1}{54}=\frac{m}{n}$ So $m+n=55$ . Not different .

The condition $x^2 + y^2 + z^2 = 1$ reminded me the sphere in $R^3$ , so I used spherical coordinates, i.e. $x=\sin\theta\sin\phi\;,\\ y=\sin\theta\cos\phi\;,\\ z=\cos\theta\;.$ The condition $x + y + z = 0$ then implies $\tan\theta(\cos\phi + \sin\phi) = -1\;,$ where I already ruled out the possibility of $\cos\theta = 0$ , i.e. $z = 0$ , since then $x^2y^2z^2 = 0$ . After some manipulation we can write $x^2y^2z^2 = \frac{1+\sin(2\phi)}{4[2 + \sin(2\phi)]^3}\sin^2(2\phi)\;.$ We now consider $-1\le\sin(2\phi)\le1$ as our new variable. The above function of $\sin(2\phi)$ has positive derivative for $-1\le\sin(2\phi)\le1$ , therefore the maximum value is attained for $\sin(2\phi) = 1$ . Thus, we have: $x^2y^2z^2 = \frac{1}{54} = \frac{a}{b}\;,$ and finally $a + b = \boxed{55}$ .