Question

Question

Answer 1

if \(\large x,y,z \in \mathbb{R^{>0}}\)

for what ratio of \(\large y\) and \(\large z\) is the value of

\(\large \color{black}{\begin{align} \left(\dfrac{x}{y}+\dfrac{z}{12x}+\dfrac{4y}{x}+\dfrac{x}{3z}\right)\hspace{.33em}\\~\\
\end{align}}\)

minimum ?

Answer 2

is the answer 1 ?

Answer 3

\(\large \color{black}{\begin{align} \dfrac{y}{z}=\dfrac14\hspace{.33em}\\~\\
\end{align}}\)

Answer 4

*

Answer 5

hint: equality of AM-GM occures when\[\large \color{black}{\begin{align} \dfrac{x}{y}=\dfrac{z}{12x}=\dfrac{4y}{x}=\dfrac{x}{3z}\hspace{.33em}\\~\\
\end{align}}\]

Answer 6

i didnt understand

Answer 7

For finding minimum of that expression you must use AM-GM and in the AM-GM minimum occurs when all of numbers are equal, In other words for positive numbers \(a_1\), \(a_2\), ... and \(a_n\) we have:\[\frac{a_1+a_2+...+a_n}{n} \ge (a_1 a_2...a_n)^{1/n}\]minimum state or equality holds if and only if\[a_1=a_2=...=a_n\]

Answer 8

Ahh nice xD
so @mukushla is \(\large 4*\sqrt[4]{9}\) the minimum value of given expression ?

Answer 9

oh yeah

Answer 10

brilliant!!

Answer 11

hey, thanks, math am I clear?

Answer 12

but

\(\large \color{black}{\begin{align} & \dfrac{x}{y}=\dfrac{z}{12x}=\dfrac{4y}{x}=\dfrac{x}{3z} \hspace{1.5em}\\~\\
& \implies \dfrac{x}{y}=\dfrac{x}{3z} \hspace{1.5em}\\~\\
& \implies \dfrac{y}{z}=\dfrac{3}{1} \hspace{1.5em}\\~\\
\end{align}}\)

Answer 13

Jesus! what happened? I found \(z=2x\) and \( y=\frac{1}{2} x\)

Answer 14

is 3/1 correct and the given 1/4 wrong ?

Answer 15

hey, I'm a little bit confused, what you did was right, but from mine:\[\frac{x}{y}=\frac{4y}{x}\]which gives \(y=\frac{1}{2}x\)\[\frac{z}{12x}=\frac{x}{3z} \]\[z=2x\]so\[\frac{y}{z}=\frac {1}{4} \]

Answer 16

I think both of them can be, because you have 3 unknowns and a lot of equations, problem is not right.

Answer 17

ok so question is incorrect ?

Answer 18

I think it is

Answer 19

because we got\[\frac{y}{z}=\frac{1}{4}=3\]which is a contradiction

Answer 20

u equated all four terms

i think it should be this only


\(\large \color{black}{\begin{align} \dfrac{x}{y}=\dfrac{4y}{x}, \dfrac{x}{3z}=\dfrac{z}{12x} \hspace{1.5em}\\~\\
\end{align}}\)
beacuse the AM and GM inequality will hold for that as they are reciprocals

Answer 21

@mukushla

Answer 22

oh no! all of them must be equal :-)

Answer 23

\(\large \color{black}{\begin{align} \dfrac{\dfrac{x}{y}+\dfrac{4y}{x}}{2}\geq \sqrt{\dfrac{x}{y}\times \dfrac{4y}{x}} \hspace{1.5em}\\~\\
\end{align}}\)

Answer 24

D:

Answer 25

sry, I gotta go, we'll talk about this later ;-)

Answer 26

ok, math, what was your reasoning?

Answer 27

these questions are so confusing...

Answer 28

i think it should be this

\(\large \color{black}{\begin{align} \dfrac{x}{y}=\dfrac{4y}{x}, \dfrac{x}{3z}=\dfrac{z}{12x} \hspace{1.5em}\\~\\
\end{align}}\) ?

Answer 29

because we can apply now
AM and GM equality now for theese for finding minimum

\(\large \color{black}{\begin{align} \dfrac{\dfrac{x}{y}+\dfrac{4y}{x}}{2}\geq \sqrt{\dfrac{x}{y}\times \dfrac{4y}{x}} \hspace{1.5em}\\~\\
\end{align}}\)

\(\large \color{black}{\begin{align} \dfrac{\dfrac{x}{3x}+\dfrac{z}{12x}}{2}\geq \sqrt{\dfrac{x}{3z}\times \dfrac{z}{12x}} \hspace{1.5em}\\~\\
\end{align}}\)

Answer 30

those to inequalities does not guarantee that \[\large \color{black}{\begin{align} \dfrac{x}{y}+\dfrac{z}{12x}+\dfrac{4y}{x}+\dfrac{x}{3z}\hspace{.33em}\\~\\
\end{align}}\]will bw minimum

Answer 31

but wolfram gives this

which says



\(x=2y,z=2x,\implies \dfrac yz=\dfrac14 \)



http://www.wolframalpha.com/input/?i=minimize+%5Cdfrac%7Bx%7D%7By%7D%2B%5Cdfrac%7Bz%7D%7B12x%7D%2B%5Cdfrac%7B4y%7D%7Bx%7D%2B%5Cdfrac%7Bx%7D%7B3z%7D%2Cx%3E0%2Cy%3E0%2Cz%3E0

Answer 32

My god, what's the matter with wolfram today?

Answer 33

AM-GM gives the minimum as\[\frac{4\sqrt{3}}{3}\]we must find the problem here!

Answer 34

I think you are right math, maybe we can't use AM-GM at all, because that four fractions can not be equal. They can't be same simultaneously.

Answer 35

What I learned here was that we must be careful when using AM-GM and actually I still have some doubts.

Answer 36

what do you think guys?

Answer 37

we apply AM GM when the variable is only one usually at numerator,
here it has 2 variables such as x/y ,y/z in numerator as well as denominator

Answer 38

@makushla , I think your strategy is valid, the problem is that the terms are not independent. To resolve, let's assign some variables first
$$
x/y+z/12x+4y/x+x/3z\\
=a+b/12+4/a+1/3b
$$
Where
a=x/y
b=z/x
Now we equate the \(dependent\) terms following your strategy:
$$
4/a=a\\
\implies a=\pm2\\
b/12=1/3b\\
\implies b=\pm 2
$$
Since we are using positive reals, we keep only positive a and b.
This means that
$$
x/y=2\\
x=2y\\
z/x=2\\
z=2x=4y\\
\implies y/z=1/4
$$

How does this sound?

Answer 39

BTW, justification for equating terms can be found here - https://en.wikipedia.org/wiki/Geometric_mean#Calculation

Answer 40

Thanks @ybarrap

Answer 41

You're welcome :)