The answer is not 6

Algebra Level 5

Let $x, y, z$ be positive real numbers such that $x^2+y^2+z^2=2(xy+yz+zx)$ . Find the minimum value of

$\frac { x+y }{ z } +\frac { y+z }{ x } +\frac { z+x }{ y } .$

Write your answer to 3 decimal places.

The answer is 10.500.

1 solution

Steven Jim
Jun 8, 2017

Assume that $x\ge y\ge z$

$\begin{aligned} \Rightarrow { x }^{ 2 }+{ (y+z) }^{ 2 } &= 2(xy+xz)+4yz \\ { (x-y-z) }^{ 2 } &= 4yz \end{aligned}$

Case 1: $x=y+z+2\sqrt { yz }$

$\Rightarrow \sqrt { x } =\sqrt { y } +\sqrt { z }$

Case 2: $y+z=x+2\sqrt { yz }$

${ (\sqrt { y } -\sqrt { z } ) }^{ 2 }=x$ $\begin{aligned} \Rightarrow \sqrt{ z } &= \sqrt{ y } + \sqrt { x } \text{ or} \\ \Rightarrow \sqrt{ y } &= \sqrt{ x } + \sqrt { z } \end{aligned}$

But $\sqrt { x }\ge \sqrt { y }\ge \sqrt { z }$ and $x, y, z$ can't be 0, so it is only possible that $\sqrt { x } =\sqrt { y } +\sqrt { z }$ .

We have

$\begin{aligned} \frac { x+y }{ z } +\frac { y+z }{ x } +\frac { z+x }{ y } &= \frac { z+y }{ { (\sqrt { z } +\sqrt { y } ) }^{ 2 } } +\frac { { { (\sqrt { z } +\sqrt { y } ) }^{ 2 } }+y }{ z } +\frac { { { (\sqrt { z } +\sqrt { y } ) }^{ 2 } }+z }{ y } \\ &\ge \frac { 1 }{ 2 } +\frac { 2y+z+2\sqrt { zy } }{ z } +\frac { 2z+y+2\sqrt { zy } }{ y } \\ &= \frac { 1 }{ 2 } +2+2(\frac { z }{ y } +\frac { y }{ z } )+2(\frac { \sqrt { z } }{ \sqrt { y } } +\frac { \sqrt { y } }{ \sqrt { z } } ) \\ &\ge 10.5 \end{aligned}$

You said that $\sqrt{x}=\sqrt{y}+\sqrt{z}$ but why compute $z=(\sqrt{x}+\sqrt{y})^2$ ? Although it doesn't matter...

Kelvin Hong - 4 years ago

Lol, I forgot :) I'll fix immediately. Thanks.

Steven Jim - 4 years ago

Very interesting question.

Some pointers.

"non-negative" includes 0, so I've changed the word to "positive". (Of course, this follows directly from the fractions, but it is better to be upfront about it.
As mentioned by Kelvin Hong, the substitution at the end should be changed.
[Not valid. I was confused by the substitution.] Note that the minimum cannot be achieved since that would require $x= y , z = 0$ . As such, I have updated the word to infimum .
The better way of presenting the start is to say that $a^4+b^4+c^4 - 2(a^2b^2+b^2c^2+c^2a^2) = (a+b+c)(a+b-c)(a-b+c)(-a+b+c)$ . This is a "well-known" factorization that appears in heron's formula .

Calvin Lin Staff - 4 years ago

Sir, so if use $\sqrt{x}=\sqrt{y} + \sqrt{z}$ correctly, we can continue doing that and state that minimum reached at y=z?

Kelvin Hong - 4 years ago

Ah haha. I got mixed up with the wrong substitution.

You are right to say that equality is achieved at (say) $x = 4, y = 1, z = 1$ . Let me revert point 3.

Calvin Lin Staff - 4 years ago

@Calvin Lin – Haha ,nevermine~ Thanks for explanation~

Kelvin Hong - 4 years ago

About number 4, I know. That's the better way, but I think that not many people know they can get $\sqrt { x } =\sqrt { y } +\sqrt { z }$ from $x^2+y^2+z^2=2(xy+yz+zx)$ without using Heron's formula (for me, I am not allowed to use this until 10th grade), so I want to post this problem and also this for people to understand it.

Steven Jim - 4 years ago

Besides, isn't it $a^4+b^4+c^4 - 2(a^2b^2+b^2c^2+c^2a^2) = (a+b+c)(a+b-c)(a-b+c)(a-b-c)$ ?

Steven Jim - 4 years ago

(=\frac { z+y }{ { (\sqrt { z } +\sqrt { y } ) }^{ 2 } } is greater than half ? How

Ayush Sharma - 3 years, 12 months ago

Just AM - GM.

Steven Jim - 3 years, 12 months ago

Actually I'm getting the answer as $-3$ . Let me explain it... Please tell whether it is correct or not. To reduce typing as I'm not too good at latex, I'll skip some steps...please forgive me for that...but I got the answer by just mere simplification.

I write the given data as $(x+y+z)^2=4(xy+yz+zx)$

So this must mean that $(x+y+z)>0$ Let this be $(i)$ and $(xy+yz+zx)>0$ , let this be $(ii)$ because they are all positive real numbers. I'll keep this aside for the time being.

Now If we simplify the expression for which min value is asked, we get, $\dfrac{x^{2}y+xy^{2}+y^{2}z+z^{2}y+z^{2}x+zx^{2}}{xyz}$

Again we'll go back... Let me multiply $(i)$ and $(ii)$ . Hence I'll be getting $x^{2}y+xy^{2}+y^{2}z+z^{2}y+z^{2}x+zx^{2}+3xyz>0$ .

Dividing throughout by $xyz$ , we get,

$\dfrac{x^{2}y+xy^{2}+y^{2}z+z^{2}y+z^{2}x+zx^{2}}{xyz}+3>0$

Hence, $\dfrac{x^{2}y+xy^{2}+y^{2}z+z^{2}y+z^{2}x+zx^{2}}{xyz}>-3$

Which is the required answer. Anything wrong in this? @Calvin Lin , can you please check it out?