The minimum value of $\Large \frac{(x^4+1)(y^4+1)(z^4+1)}{xy^2z}$ as x,y and z range over the positive reals is equal to $\Large \frac{A\sqrt{B}}{C}$, where A and C are coprime and B is squarefree. What is $\Large A+B+C\ ?$

Question

vishweshshrimali5 · Answer

We would have to use inequalities

anonymous · Answer

yes

vishweshshrimali5 · Answer

Any idea ?

anonymous · Answer

Using AM-GM

vishweshshrimali5 · Answer

Have you tried that ?

vishweshshrimali5 · Answer

Gotcha !

vishweshshrimali5 · Answer

Just a minute

vishweshshrimali5 · Answer

I can write the given expression as:

$$\Large {(x^3+\cfrac{1}{x})(y^2+\cfrac{1}{y^2})(z^3+\cfrac{1}{z})}$$

vishweshshrimali5 · Answer

Now I can apply AM - GM to the second one

vishweshshrimali5 · Answer

For first and third, I would have to think something...

anonymous · Answer

$\Large x^3+\frac{1}{x} =x^3+\frac{1}{3x}+\frac{1}{3x}+\frac{1}{3x}$
and using AM-GM

vishweshshrimali5 · Answer

Yup perfect

vishweshshrimali5 · Answer

If you use AM - GM to all three, then it would simplify to :

$$\large{\cfrac{32\sqrt{3}}{9}}$$

vishweshshrimali5 · Answer

Co prime - check
square free - check

vishweshshrimali5 · Answer

A+B+C = 44

vishweshshrimali5 · Answer

Done !

vishweshshrimali5 · Answer

Another method - calculus !

vishweshshrimali5 · Answer

Have you studied calculus @JungHyunRan ?

anonymous · Answer

yes

vishweshshrimali5 · Answer

Great!

vishweshshrimali5 · Answer

Then using partial differentiation and maxima - minima, you can come up with the same answer more easily

anonymous · Answer

^^ I see. Thank you so much @vishweshshrimali5

vishweshshrimali5 · Answer

No problem and it was entertaining to solve such problems after a very long time :)

vishweshshrimali5 · Answer

Remember such problems which involve symmetry (as in here), calculus provides the shortest answers

ganeshie8 · Answer

I had seen AM-GM abused this nicely a couple of days back... for the first time, by @satellite73  :)

vishweshshrimali5 · Answer

:D @satellite73 ROCKS !!

vishweshshrimali5 · Answer

He has not got a 100 smart score for nothing after all !

vishweshshrimali5 · Answer

@JungHyunRan, if you like such type of problems I would suggest you to go through some olympiad level books. They abuse very very simple inequalities in interesting ways :)

anonymous · Answer

yup

vishweshshrimali5 · Answer

A great collection of books can be found at this link: http://www.imomath.com/index.php?options=347&lmm=0

ganeshie8 · Answer

im still not able to get to that problem solved by satellite, my answered questions are not loading >.< will post it when i find it :)

vishweshshrimali5 · Answer

Sure @ganeshie8 :)