Maximum and minimum value of 2xy+2yz+7xz2xy+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,zx, y, z be real numbers such that x2+y2+z2100x^2+y^2+z^2\leq 100, find the maximum and minimum values of 2xy+2yz+7xz.2xy+2yz+7xz..

Note by Jorge Tipe
5 years, 9 months ago

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I hate to do this, but I feel inclined to bash using Lagrange multipliers. For any x2+y2+z2=kx^2+y^2+z^2=k, 0k1000\le k\le100, we want <2y+7z,2x+2z,2y+7x>=λ<2x,2y,2z>\left<2y+7z,2x+2z,2y+7x\right>=\lambda\left<2x,2y,2z\right>. Solving for λ\lambda, we get 2y+7z=2xλ2y+7z=2x\lambda, 2x+2z=2yλ2x+2z=2y\lambda, and 2y+7x=2zλ2y+7x=2z\lambda. Eliminate xx by substituting x=yλzx=y\lambda-z to get 2y+7z=2yλ22zλ2y+7z=2y\lambda^2-2z\lambda and 2y+7(yλz)=2zλ2y+7(y\lambda-z)=2z\lambda. Plugging in the second as 2zλ2z\lambda, we get the quadratic 0=2yλ27yλ4y=y(2λ+1)(λ4)0=2y\lambda^2-7y\lambda-4y=y(2\lambda+1)(\lambda-4), so λ=12 or 4\lambda=-\frac{1}{2}\text{ or }4, or y=0y=0. Whew, λ\lambda is somewhat easy to work with (thanks Jorge!).

In the case that y=0y=0, we have 7xz7xz to optimize. Assume WLOG that x,z0x,z\ge0, so 7xz7(x2+z22)271002=3507xz\le7\left(\sqrt{\frac{x^2+z^2}{2}}\right)^2\le7\frac{100}{2}=350. That means y=0y=0 gives a maximum of 350350 (x=z=52x=z=5\sqrt{2}) and minimum of 350-350 (x=52,z=52x=5\sqrt{2},z=-5\sqrt{2}).

In the case that λ=4\lambda=4, 2y+7z=4(2x)2y+7z=4(2x) and 2x+2z=4(2y)2x+2z=4(2y). Solving for yy and zz, y=12xy=\frac{1}{2}x and z=xz=x, which plugs in as 2xy+2yz+5zx=9x22xy+2yz+5zx=9x^2 and x2+y2+z2=94x2100x^2+y^2+z^2=\frac{9}{4}x^2\le100. The minimum here is 00 (x=y=z=0x=y=z=0) clearly, but the maximum is 949100=4009\frac{4}{9}100=400 (x=z=203,y=103x=z=\frac{20}{3},y=\frac{10}{3}).

In the case that λ=12\lambda=-\frac{1}{2}, 2y+7z=12(2x)2y+7z=-\frac{1}{2}(2x) and 2x+2z=12(2y)2x+2z=-\frac{1}{2}(2y). Solving for yy and zz, we get y=4xy=-4x and z=xz=x, which plugs in as 2xy+2yz+5zx=11x22xy+2yz+5zx=-11x^2 and x2+y2+z2=18x2100x^2+y^2+z^2=18x^2\le100. The maximum here is clearly 00 (x=y=z=0x=y=z=0), but the minimum is 410018=5509-4\frac{100}{18}=-\frac{550}{9} (x=z=523,y=2023x=z=\frac{5\sqrt{2}}{3},y=-\frac{20\sqrt{2}}{3}).

In effect, the maximum value is 400\boxed{400} at (203,103,203)\left(\frac{20}{3},\frac{10}{3},\frac{20}{3}\right), and the minimum value is 350\boxed{-350} at (52,0,52)\left(5\sqrt{2},0,-5\sqrt{2}\right).

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

Cody Johnson - 5 years, 9 months ago

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I was reading your solution very carefully trying to spot the mistake because your answers were wrong, then I realized that you edited while I was reading... now your answers are correct!

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.

Jorge Tipe - 5 years, 9 months ago

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Is it Jensen's?

Cody Johnson - 5 years, 9 months ago

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

A Former Brilliant Member - 5 years, 2 months ago

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If we want the greatest possible value of 2xy+2yz+2xz2xy+2yz+2xz (instead of 2xy+2yz+7xz2xy+2yz+7xz ) we can use the inequalities 2xyx2+y22xy\leq x^2+y^2, 2yzy2+z22yz\leq y^2+z^2 and 2xzx2+z22xz\leq x^2+z^2, in this case the maximum occurs when x=y=zx=y=z.

But in our problem we can try to use similar inequalities like 2xyAx2+By22xy\leq Ax^2+By^2, where AA and BB are suitable constants.

Jorge Tipe - 5 years, 9 months ago

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For minimum: 2xy+2yz+7xz=(x+y+z)2(x2+y2+z2)+52(x+z)252(x2+z2)(x2+y2+z2)52(x2+z2)10052(100)=350. 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,x2+y2+z2=100,x2+z2=100x+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,x2=z2=50y=0, x=-z, x^2=z^2=50.

George G - 5 years, 9 months ago

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

First note that xx and zz have a symmetric role. Then, to obtain x2+y2+z2x^2+y^2+z^2 I will work with a expression of the form (x+λy+z)2(x+\lambda y+z)^2 [note that xx and zz have the same coefficient]]. If we want a multiple of 2xy+2yz+7xz2xy+2yz+7xz we can take λ=27\lambda=\frac{2}{7}.

Thus we work with the following inequality which is true: (x+27y+z)2+4549y20,\left(x+\frac{2}{7}y+z\right)^2+\frac{45}{49}y^2 \geq 0, this inequality is equivalent to: 27(2xy+2yz+7xz)x2y2z2\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]

Jorge Tipe - 5 years, 9 months ago

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Very nice, interesting approach.

George G - 5 years, 9 months ago

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

Eddie The Head - 5 years, 9 months ago

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Really nice approach.

Eddie The Head - 5 years, 9 months ago

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2xy+2yz+7xz=what?

jagadesh kumar - 5 years, 9 months ago

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Let k=2xy+2yz+7zxk = 2xy+2yz+ 7zx. Then k100(2xy+2yz+7zx)x2+y2+z2k \leq \dfrac{100( 2xy+2yz+ 7zx)}{x^2+y^2+z^2}, which becomes kx2x(200y+700z)+ky2200yz+kz20.\begin{aligned} kx^2-x(200y+700z)+ky^2-200yz+kz^2\leq 0. \end{aligned} Since the solutions of kx2x(200y+700z)+ky2200yz+kz2=0kx^2-x(200y+700z)+ky^2-200yz+kz^2 = 0 are real (solutions are xx and 200y+700zkx\dfrac{200y+700z}{k}-x), its discriminant Δx\Delta_x is nonnegative: Δx=(200y+700z)24k(ky2200yz+kz2)0    (400004k2)y2+(280000+800k)yz+(4900004k2)z20.()\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=0z = 0, then ()(\star) becomes (400004k2)y20(40000-4k^2)y^2 \geq 0 and since y20y^2\geq 0 for all yRy\in\mathbb{R} we must have 400004k20    k10040000-4k^2\geq 0\implies |k|\leq 100. If z0z\neq 0, then dividing both sides of ()(\star) and taking t=yzt=\dfrac{y}{z} gives (400004k2)t2+(280000+800k)t+(4900004k2)0.\begin{aligned} (40000-4k^2)t^2 + (280000+800k)t+(490000-4k^2)\geq 0. \end{aligned} Since the roots of the equation (400004k2)t2+(280000+800k)t+(4900004k2)=0(40000-4k^2)t^2 + (280000+800k)t+(490000-4k^2)=0 are real (note that t=yzRt = \dfrac{y}{z}\in\mathbb{R}, its discriminant Δt\Delta_t is nonnegative: Δt=(280000+800k)24(400004k2)(4900004k2)0    k3142500k70000000.\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 k3142500k7000000=0k^3-142500k-7000000=0 are k=350,50,400k=-350, -50, 400 (also obtained from Wolfram Alpha), the minimum value of k=2xy+2yz+7zxk = 2xy+2yz+7zx is 350\boxed{-350} at (x,y,z)=(52,0,52)(x,y,z)=(-5\sqrt{2}, 0, 5\sqrt{2}) and the maximum value is 400\boxed{400} at (x,y,z)=(203,103,203)(x,y,z)=\left(\dfrac{20}{3}, \dfrac{10}{3}, \dfrac{20}{3}\right).

Russelle Guadalupe - 5 years, 2 months ago

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

Cody Johnson - 5 years, 2 months ago

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For maximum: 2xy+2yz+7xz=4(x2+y2+z2)12((x2y)2+(z2y)2+7(xz)2)400. 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=zx=2y=z and x2+y2+z2=100x^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[x2+(2y)2+x2+x2+x2+z2+z2+z2+z2][(2y)2+z2+z2+z2+z2+x2+x2+x2+x2]=(4x2+4y2+4z2)2 (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)12(x2+4y2),2yz=(2y)z12(z2+4y2),7xz72(x2+z2) 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.

George G - 5 years, 9 months ago

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Why has the diagram of a sphere been given?

Rohit Nair - 5 years, 9 months ago

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It's only a point inside a sphere :) If x2+y2+z2100x^2+y^2+z^2\leq 100 then the point (x,y,z)(x,y,z) is inside the sphere of radius 10.

Jorge Tipe - 5 years, 9 months ago

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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

x2+y2+z23[(xy)(yz)(zx)]13xy+yz+zx3\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,

x2+y2+z2(x+y+z)23x^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)+5zx2xy+2yz+7zx=2(xy+yz+zx)+5zx

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

Rahul Saha - 5 years, 9 months ago

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I think something like 2xy+2yz+7zx(22+22+72)(x2y2+y2z2+z2x2)=57(x2y2+y2z2+z2x2)57(x4+y4+z4)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 x2y2x^2y^2 (and similar terms) are nonnegative.

Arkan Megraoui - 5 years, 9 months ago

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That gives you an upper bound on the possible value of 2xy+2yz+7zx 2xy+2yz+7zx . Another (easy) upper bound is that since x,y,z10 |x|, |y|, |z| \leq 10 , hence the value is at most 200+200+700=1100 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?

Calvin Lin Staff - 5 years, 9 months ago

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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,zx,y,z is positive?

Jit Ganguly - 5 years, 9 months ago

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No,but x2,y2,z2x^2,y^2,z^2 are always non-negative.However,when I say

[(xy)(yz)(zx)]13xy+yz+zx3 [(xy)(yz)(zx)]^\dfrac{1}{3}\le \dfrac{xy+yz+zx}{3}

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

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

Rahul Saha - 5 years, 9 months ago

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@Rahul Saha I see! Thanks Rahul!

Jit Ganguly - 5 years, 9 months ago

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how can i become as intelligent as you

Atul Tiwari - 5 years, 9 months ago

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

Cody Johnson - 5 years, 9 months ago

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

Ahaan Rungta - 5 years, 9 months ago

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@Ahaan Rungta 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
Read the Techniques page. etc.

Calvin Lin Staff - 5 years, 9 months ago

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How can you be so rude??

Anuj Shikarkhane - 5 years, 1 month ago

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