# Maximum and minimum value of $2xy+2yz+7xz$

This is a problem I proposed for the Final round of the Peruvian Mathematical Olympiad 2013, about 1 month ago. I hope you will enjoy this problem.

Problem Let $x, y, z$ be real numbers such that $x^2+y^2+z^2\leq 100$, find the maximum and minimum values of $2xy+2yz+7xz.$.

Note by Jorge Tipe
7 years ago

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Let $k = 2xy+2yz+ 7zx$. Then $k \leq \dfrac{100( 2xy+2yz+ 7zx)}{x^2+y^2+z^2}$, which becomes \begin{aligned} kx^2-x(200y+700z)+ky^2-200yz+kz^2\leq 0. \end{aligned} Since the solutions of $kx^2-x(200y+700z)+ky^2-200yz+kz^2 = 0$ are real (solutions are $x$ and $\dfrac{200y+700z}{k}-x$), its discriminant $\Delta_x$ is nonnegative: \begin{aligned} \Delta_x = (200y+700z)^2-4k(ky^2-200yz+kz^2) &\geq 0\\ \implies (40000-4k^2)y^2 + (280000+800k)yz+(490000-4k^2)z^2&\geq 0.\qquad(\star) \end{aligned} If $z = 0$, then $(\star)$ becomes $(40000-4k^2)y^2 \geq 0$ and since $y^2\geq 0$ for all $y\in\mathbb{R}$ we must have $40000-4k^2\geq 0\implies |k|\leq 100$. If $z\neq 0$, then dividing both sides of $(\star)$ and taking $t=\dfrac{y}{z}$ gives \begin{aligned} (40000-4k^2)t^2 + (280000+800k)t+(490000-4k^2)\geq 0. \end{aligned} Since the roots of the equation $(40000-4k^2)t^2 + (280000+800k)t+(490000-4k^2)=0$ are real (note that $t = \dfrac{y}{z}\in\mathbb{R}$, its discriminant $\Delta_t$ is nonnegative: \begin{aligned} \Delta_t = (280000+800k)^2-4(40000-4k^2)(490000-4k^2)\geq 0\\ \implies k^3-142500k-7000000\leq 0. \end{aligned} (I used Wolfram Alpha to obtain this cubic inequality). Since the roots of $k^3-142500k-7000000=0$ are $k=-350, -50, 400$ (also obtained from Wolfram Alpha), the minimum value of $k = 2xy+2yz+7zx$ is $\boxed{-350}$ at $(x,y,z)=(-5\sqrt{2}, 0, 5\sqrt{2})$ and the maximum value is $\boxed{400}$ at $(x,y,z)=\left(\dfrac{20}{3}, \dfrac{10}{3}, \dfrac{20}{3}\right)$.

- 6 years, 5 months ago

Woah! Amazing solution! Truly genius, in my opinion!

- 6 years, 5 months ago

For minimum: $2xy+2yz+7xz = (x+y+z)^2 - (x^2+y^2+z^2) + \frac{5}{2} (x+z)^2 - \frac{5}{2}(x ^2 + z^2) \\ \ge - (x^2+y^2+z^2) - \frac{5}{2}(x ^2 + z^2) \ge -100 - \frac{5}{2} (100) = -350.$ The equality holds when $x+y+z=0, x+z=0, x^2+y^2+z^2=100, x^2+z^2=100$. There is an obvious solution, that is $y=0, x=-z, x^2=z^2=50$.

- 7 years ago

What was your motivation behind breaking it up like this?....

- 7 years ago

Really nice approach.

- 7 years ago

2xy+2yz+7xz=what?

- 7 years ago

For me is very nice to see new solutions of this problem. Let me show my approach:

First note that $x$ and $z$ have a symmetric role. Then, to obtain $x^2+y^2+z^2$ I will work with a expression of the form $(x+\lambda y+z)^2$ [note that $x$ and $z$ have the same coefficient]]. If we want a multiple of $2xy+2yz+7xz$ we can take $\lambda=\frac{2}{7}$.

Thus we work with the following inequality which is true: $\left(x+\frac{2}{7}y+z\right)^2+\frac{45}{49}y^2 \geq 0,$ this inequality is equivalent to: $\frac{2}{7}(2xy+2yz+7xz)\geq -x^2-y^2-z^2$ and we are done! [Of course, we have to complete some details]

- 7 years ago

Very nice, interesting approach.

- 7 years ago

For maximum: $2xy+2yz+7xz = 4(x^2+y^2+z^2) - \frac{1}{2}((x-2y)^2+(z-2y)^2 + 7(x-z)^2) \le 400.$ The equality holds when $x=2y=z$ and $x^2+y^2+z^2=100$.

Or apply Cauchy-Schwarz to:

$(2xy+2yz+7xz)^2 = [(x)(2y) + (2y)(z) + (x)(z) + (x)(z)+ (x)(z) + (z)(x) + (z)(x)+(z)(x)+(z)(x)]^2\\ \le[x^2+(2y)^2+x^2+x^2+x^2+z^2+z^2+z^2+z^2][(2y)^2+z^2+z^2+z^2+z^2+x^2+x^2+x^2+x^2]\\ =(4x^2+4y^2+4z^2)^2$

Or simply as Jorge pointed out: $2xy = x(2y)\le \frac{1}{2}(x^2+4y^2), 2yz = (2y)z\le \frac{1}{2}(z^2+4y^2), 7xz \le \frac{7}{2}(x^2+z^2)$ then add all three together.

- 7 years ago

how can i become as intelligent as you

- 7 years ago

Definitely not by asking "how can i become as intelligent as you."

- 7 years ago

How can you be so rude??

- 6 years, 4 months ago

If you think being sarcastic sounds clever, you're highly mistaken.

- 7 years ago

The tone of sarcasm rarely transmits well over the internet, so let's avoid it. I'm sure we can come up with several pointers, like:

Practice, practice, practice
Learn from others

Staff - 7 years ago

If we want the greatest possible value of $2xy+2yz+2xz$ (instead of $2xy+2yz+7xz$ ) we can use the inequalities $2xy\leq x^2+y^2$, $2yz\leq y^2+z^2$ and $2xz\leq x^2+z^2$, in this case the maximum occurs when $x=y=z$.

But in our problem we can try to use similar inequalities like $2xy\leq Ax^2+By^2$, where $A$ and $B$ are suitable constants.

- 7 years ago

I have no idea what to do here.I tried establishing a bound but I am not sure what to use here.I have tried to use some elementary inequalities such as Cauchy-Schwarz and AM-GM.Using AM-GM,we find that

$\dfrac{x^2+y^2+z^2}{3}\ge [(xy)(yz)(zx)]^\dfrac{1}{3}\le \dfrac{xy+yz+zx}{3}$

But I am not sure how that helps us.Also,

$x^2+y^2+z^2\ge \dfrac{(x+y+z)^2}{3}$

by Titu's lemma.I have also tried to rewrite the expression as

$2xy+2yz+7zx=2(xy+yz+zx)+5zx$

and was wondering whether I can apply AM-GM here.

- 7 years ago

I think something like $2xy+2yz+7zx\le (2^2+2^2+7^2)(x^2y^2+y^2z^2+z^2x^2)=57(x^2y^2+y^2z^2+z^2x^2)\le 57(x^4+y^4+z^4)$ would help, because $x^2y^2$ (and similar terms) are nonnegative.

- 7 years ago

That gives you an upper bound on the possible value of $2xy+2yz+7zx$. Another (easy) upper bound is that since $|x|, |y|, |z| \leq 10$, hence the value is at most $200+200+700=1100$. Hence, we also need to check and see if it is indeed the least upper bound.

For it to be the least upper bound, we must have equality hold throughout. What are the equality conditions for each of your inequalities?

Staff - 7 years ago

Hi Rahul! If I'm not mistaken for using AM. G.M the numbers on which the inequality is applied must be positive, here is there any condition that $x,y,z$ is positive?

- 7 years ago

No,but $x^2,y^2,z^2$ are always non-negative.However,when I say

$[(xy)(yz)(zx)]^\dfrac{1}{3}\le \dfrac{xy+yz+zx}{3}$

I am possibly wrong,since we don't know if $xy,yz,zx$ are non-negative.However,when looking for the maximum,we can safely assume that $x,y,z$ are positive.

NOTE: AM-GM stipulates the values being non-negative ,not positive.

- 7 years ago

I see! Thanks Rahul!

- 7 years ago

I hate to do this, but I feel inclined to bash using Lagrange multipliers. For any $x^2+y^2+z^2=k$, $0\le k\le100$, we want $\left<2y+7z,2x+2z,2y+7x\right>=\lambda\left<2x,2y,2z\right>$. Solving for $\lambda$, we get $2y+7z=2x\lambda$, $2x+2z=2y\lambda$, and $2y+7x=2z\lambda$. Eliminate $x$ by substituting $x=y\lambda-z$ to get $2y+7z=2y\lambda^2-2z\lambda$ and $2y+7(y\lambda-z)=2z\lambda$. Plugging in the second as $2z\lambda$, we get the quadratic $0=2y\lambda^2-7y\lambda-4y=y(2\lambda+1)(\lambda-4)$, so $\lambda=-\frac{1}{2}\text{ or }4$, or $y=0$. Whew, $\lambda$ is somewhat easy to work with (thanks Jorge!).

In the case that $y=0$, we have $7xz$ to optimize. Assume WLOG that $x,z\ge0$, so $7xz\le7\left(\sqrt{\frac{x^2+z^2}{2}}\right)^2\le7\frac{100}{2}=350$. That means $y=0$ gives a maximum of $350$ ($x=z=5\sqrt{2}$) and minimum of $-350$ ($x=5\sqrt{2},z=-5\sqrt{2}$).

In the case that $\lambda=4$, $2y+7z=4(2x)$ and $2x+2z=4(2y)$. Solving for $y$ and $z$, $y=\frac{1}{2}x$ and $z=x$, which plugs in as $2xy+2yz+5zx=9x^2$ and $x^2+y^2+z^2=\frac{9}{4}x^2\le100$. The minimum here is $0$ ($x=y=z=0$) clearly, but the maximum is $9\frac{4}{9}100=400$ ($x=z=\frac{20}{3},y=\frac{10}{3}$).

In the case that $\lambda=-\frac{1}{2}$, $2y+7z=-\frac{1}{2}(2x)$ and $2x+2z=-\frac{1}{2}(2y)$. Solving for $y$ and $z$, we get $y=-4x$ and $z=x$, which plugs in as $2xy+2yz+5zx=-11x^2$ and $x^2+y^2+z^2=18x^2\le100$. The maximum here is clearly $0$ ($x=y=z=0$), but the minimum is $-4\frac{100}{18}=-\frac{550}{9}$ ($x=z=\frac{5\sqrt{2}}{3},y=-\frac{20\sqrt{2}}{3}$).

In effect, the maximum value is $\boxed{400}$ at $\left(\frac{20}{3},\frac{10}{3},\frac{20}{3}\right)$, and the minimum value is $\boxed{-350}$ at $\left(5\sqrt{2},0,-5\sqrt{2}\right)$.

Now the real question is what is the non-LM solution?

- 7 years ago

This is exactly how I would've done it, great solution, regardless of any bashing

- 6 years, 5 months ago

Of course there is a elementary solution, in fact this problem was proposed to students with no knowledge of Calculus (grades 9 and 10 in Peru) and only one student managed to solve both parts of the problem.

- 7 years ago

Is it Jensen's?

- 7 years ago

Why has the diagram of a sphere been given?

- 7 years ago

It's only a point inside a sphere :) If $x^2+y^2+z^2\leq 100$ then the point $(x,y,z)$ is inside the sphere of radius 10.

- 7 years ago

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