- mathmath333

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- katieb

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- mathmath333

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 ?

- ganeshie8

is the answer 1 ?

- mathmath333

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

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## More answers

- ikram002p

*

- anonymous

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

- mathmath333

i didnt understand

- anonymous

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\[a_1=a_2=...=a_n\]

- ganeshie8

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

- anonymous

oh yeah

- ganeshie8

brilliant!!

- anonymous

hey, thanks, math am I clear?

- mathmath333

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

- anonymous

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

- mathmath333

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

- anonymous

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\[\frac{y}{z}=\frac {1}{4} \]

- anonymous

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

- mathmath333

ok so question is incorrect ?

- anonymous

I think it is

- anonymous

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

- mathmath333

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

- mathmath333

- anonymous

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

- mathmath333

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

- mathmath333

D:

- anonymous

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

- anonymous

ok, math, what was your reasoning?

- acxbox22

these questions are so confusing...

- mathmath333

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

- mathmath333

because we can apply now
AM and GM equality now for theese for finding minimum
\(\large \color{black}{\begin{align} \dfrac{\dfrac{x}{y}+\dfrac{4y}{x}}{2}\geq \sqrt{\dfrac{x}{y}\times \dfrac{4y}{x}} \hspace{1.5em}\\~\\
\end{align}}\)
\(\large \color{black}{\begin{align} \dfrac{\dfrac{x}{3x}+\dfrac{z}{12x}}{2}\geq \sqrt{\dfrac{x}{3z}\times \dfrac{z}{12x}} \hspace{1.5em}\\~\\
\end{align}}\)

- anonymous

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

- mathmath333

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

- anonymous

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

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

what do you think guys?

- mathmath333

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

- ybarrap

@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?

- ybarrap

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

- anonymous

Thanks @ybarrap

- ybarrap

You're welcome :)

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