Very Weird Inequality

Algebra Level 4

Let $0 < x \leq y \leq z$ such that $xy+yz+zx=3$ .

Find the maximum value of $xy^3z^2$ .

The answer is 2.00.

1 solution

Mark Hennings
Feb 9, 2017

Note that, for $0 < x \le y \le z$ and $xy + xz + yz = 3$ , we have $\begin{aligned} \tfrac{1}{256}(xy^3z^2)^2 & = \tfrac{1}{256}x^2y^6z^4 \; \le \; \tfrac{1}{256}x^2y^5z^5 \; = \; xy \times xz \times \big(\tfrac14yz\big)^4 \\ & \le \left(\tfrac16\big(xy + xz + 4\times\tfrac14yz\big)\right)^6 \; = \; \left(\tfrac16(xy + xz + yz)\right)^6 \; \le \; \tfrac{1}{64} \end{aligned}$ so that $xy^3z^2 \le 2$ . This inequality can be achieved when $y = z = 4x$ , so when $x=\tfrac{1}{2\sqrt{2}}$ , $y=z=\sqrt{2}$ . Thus the maximum value is $\boxed{2}$ .

Moderator note:

As mentioned in the comments, the substitution of $a = xy, b = xz, c = yz$ subject to $0 \leq a \leq b \leq c$ and $a + b + c = 3$ , makes this problem easier to tackle.

We want to maximize $xy^3 z^2 = a c^2$ , which doesn't yet have a $b$ term. We can introduce $b$ by bounding above with $a^{1-p} b^{p} c^2$ for any $0 < p < 1$ . Then, weighted AM-GM gives us

$\frac{3}{3} = \frac{ (1 - p) \times \frac{a}{1-p} + p \times \frac{b}{p} + 2 \times \frac{c}{2} } {(1-p)+p+2} \geq \sqrt[(1-p)+p+2]{ \left(\frac{a}{1-p}\right)^{1-p} \left(\frac{b}{p}\right)^p \left(\frac{c}{2}\right)^2}.$

Hence, $a c^2 \leq a^{1-p} b^p c^2 \leq 4 (1-p)^{1-p} p^p$ for all $0 < p < 1$ .

By differentiating, we see that the RHS is minimized when $p = \frac{1}{2}$ , and that is why the choice of $ac^2 \leq \sqrt{ab} c^2$ was the right one to make.

Alternatively, note that satisfying the equality condition requires $\frac{a}{1-p} = \frac{b}{p}$ . Since $a \leq b$ , this implies that $p \geq 1-p$ . Then, since $4 (1-p)^{1-p} p^p$ is increasing on $[\frac{1}{2}, 1 ]$ , we choose $p = \frac{1}{2}$ .

That's a very interesting trick! What motivated it?

I tried to do $\sqrt[3]{xy^2 z^2} \leq \frac{ xy + y z + yz } { 3}$ , but then got stuck optimizing the RHS even further.

Calvin Lin Staff - 4 years, 4 months ago

If you check out my report, I initially analysed this by a change of variable, which made it easy to solve the original, incorrect, problem. It was then easy to make the necessary change to the corrected problem. This solution is what happens if you don't make the change of variables!

Mark Hennings - 4 years, 4 months ago

This is an ingenious way of grouping the terms together. I have no idea how you managed to pull it off. Sharky, is this also your intended approach? Or do you also have a clever approach like this?

I actually thought about using calculus to find the extrema point (via Lagrange + Compact set stuffs that you mentioned last time), but the algebra just gets so tedious that I gave up half way. Sigh....

Pi Han Goh - 4 years, 3 months ago

You might find the trick easier to see if you make the substitution $a=xy,b=xz,c=yz$ . Then we are trying to maximise $ac^2$ subject to $0 \le a \le b \le c$ and $a+b+c=3$ . Then write $ac^2 \; \le \; \sqrt{ab}c^2$ and apply the AM/GM result.

Mark Hennings - 4 years, 3 months ago

I found it difficult to push through by using any other bounding, so I studied this function further. As it turns out, the rest would not work well, and so I've added more details as to why this choice was "ideal".

Calvin Lin Staff - 4 years, 3 months ago

What is wrong with this method: Let $x=\sqrt{\frac{3}{6n+2}},y=2nx,z=2x$ They satisfy the condition, and as $n$ grows large $xy^3z^2$ approaches $4$ .

Shaun Leong - 4 years, 3 months ago

$xy^3z^2 = 16n^3 x^6$ tends to $2$ as $n\to\infty$

Mark Hennings - 4 years, 3 months ago

Isn't it $32n^3x^6$ ?

Shaun Leong - 4 years, 3 months ago

@Shaun Leong – Sorry, it is. You problem is that you don't have $x \le y \le z$ . You have $y > z$ .

Mark Hennings - 4 years, 3 months ago

@Mark Hennings – Oops thanks

Shaun Leong - 4 years, 3 months ago

The problem can be solved by means of calculus. We consider dealing with the conditional extremum, and form Lagrange's function $f(x, y, z)=x y^3 z^2 + k(x y + y z + z x-3)$ . We solve the system $f_x=y^3z^2+ky+kz=0,\quad f_y=3xy^2z^2+kx+kz=0,\quad f_z=2xy^3z+ky+kx=0,\quad xy+yz+zx=3,$ by multiplying the first equation by $x$ , the second by $y$ , and the third by $z$ . Using the fourth equation, we obtain $xy^3z^2+k(3-yz)=0,\quad 3xy^3z^2+k(3-xz)=0,\quad 2xy^3z^2+k(3-xy)=0,\quad xy+yz+zx=3.$ Summing the first three equations, we have $xy^3z^2=-k$ . Replacing $y^3z^2=-\dfrac kx$ in the first equation and using the fourth equation, we find $y z = 2, x =\dfrac1{y + z}$ , and afterwards $y = z$ . So, $y = z = \sqrt 2$ and $x = \dfrac1{2\sqrt 2}$ , i.e. the required maximal value is $2$ . We ascertain that $2$ is the maximal value, because $0<x y^3 z^2=4xy\,(yz)^2=\frac{4y}{y+z}\leq\frac{4y}{2y}=2,$ since $y\leq z$ . So, for $y = z = \sqrt 2$ and $x = \dfrac1{2\sqrt 2}$ the function $x y^3 z^2$ reaches the maximal value $2$ .

A Former Brilliant Member - 3 years, 6 months ago

How do you know it's a maximum value? Maybe could be a saddle point?

Pi Han Goh - 3 years, 6 months ago

Very Weird Inequality

The answer is 2.00.

1 solution

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