Why isn't it B cubed?

Algebra Level 4

$\dfrac{1}{a^4}+\dfrac{1}{b^4}+\dfrac{1}{c^4}=1$

Given the above equation for positive numbers $a,b,c$ .

Find the minimum value of

$\dfrac{a^4b^4+a^4c^4+b^4c^4}{a^3b^2c^3}$

If the minimum value of the above is $x$ , input your answer as $\lfloor 100x \rfloor$ .

This is part of the set Trevor's Ten

Details and Assumptions

The answer is not $300$ .
It is indeed $a^3 b^2 c^3$ and not $a^3 b^3 c^3$

The answer is 282.

4 solutions

Trevor Arashiro
Mar 15, 2015

Solution

Simplifying our semi gigantic fraction.

$\dfrac{a^4b^4+a^4c^4+b^4c^4}{a^3b^2c^3}=\dfrac{ab^2}{c^3}+\dfrac{ac }{b^2}+\dfrac{b^2c}{a^3}$

$\dfrac{ab^2}{c^3}+\dfrac{ac }{b^2}+\dfrac{b^2c}{a^3}=ab^2c\left(\dfrac{1}{a^4}+\dfrac{1}{b^4}+\dfrac{1}{c^4}\right)$

Remember that $\dfrac{1}{a^4}+\dfrac{1}{b^4}+\dfrac{1}{c^4}=1$ . Thus the above can be simplified to $ab^2c$ .

Thus we are simply trying to minimize $ab^2c$ . By G.M. $\leq$ A.M.

$\sqrt[4]{\dfrac{1}{4a^4b^8c^4}}\leq \dfrac{\dfrac{1}{a^4}+\dfrac{1}{2b^4}+\dfrac{1}{2b^4}+\dfrac{1}{c^4}}{4}$

Simple algrbra yields.

$2\sqrt{2}\leq ab^2c$

Thus our answer is

$\lfloor 2\sqrt2 \cdot100\rfloor=\boxed{282}$

I did it in a basically equivalent way:

Note that the condition is $a^4b^4+b^4c^4+c^4a^4=a^4b^4c^4$ so we just want to find the minimum of $\dfrac{a^4b^4c^4}{a^3b^2c^3}=ab^2c$

Now by GM-HM inequality, $\sqrt[4]{4a^4b^8c^4}\ge \dfrac{4}{\dfrac{1}{a^4}+\dfrac{1}{2b^4}+\dfrac{1}{2b^4}+\dfrac{1}{c^4}}=4$ so $ab^2c\ge \dfrac{4}{\sqrt[4]{4}}=2\sqrt{2}$

Daniel Liu - 6 years, 3 months ago

Wow, nice alternative. So many solutions, all relatively the same but all with their own unique twist.

Trevor Arashiro - 6 years, 3 months ago

I did the same

Department 8 - 5 years, 6 months ago

i did same!

Dev Sharma - 5 years, 5 months ago

Why is this rated as level 5 with 350 points? Is it that hard? Also, why did it get an official rating after I solved it. Why am I so unlucky?

Prasun Biswas - 6 years, 3 months ago

I don't know what to mention now , since my method has been pretty much covered with what Daniel and Prasun have posted .

Btw your question is indeed genuine , why isn't it $B^{3}$ ?

@Calvin Lin sir , is there any reason for it ? Sorry for the trouble though .

A Former Brilliant Member - 6 years, 3 months ago

I'm not sure what u mean. U mean brilliant cubed? Lol

Trevor Arashiro - 6 years, 3 months ago

Yes , why did the staff choose B^2 ! Just curious now that you have asked it as a question

A Former Brilliant Member - 6 years, 3 months ago

@A Former Brilliant Member – Let $B$ denote the actions we make on this site. Then $B^2$ can be considered as the effect it produces on the community. Now, there are people who produce negative actions on this site (example: people like me). But as you can see, $B^2$ always remains non-negative. That is the reason.

Prasun Biswas - 6 years, 3 months ago

@Prasun Biswas – Yeah , the smallest even natural number with that property . Nice explanation :)

A Former Brilliant Member - 6 years, 3 months ago

Is the given condition necessary? I solved it like this which didn't really need the given condition. Here it goes:

The given expression (say $(X)$ ) simplifies to,

$X=\left(\frac{ab^2}{c^3}+\frac{ac}{b^2}+\frac{b^2c}{a^3}\right)=b^2\left(\frac{a}{c^3}+\frac{c}{a^3}\right)+\frac{ac}{b^2}$

Applying AM-GM, we have,

$\frac{a}{c^3}+\frac{c}{a^3}\geq \frac{2}{ac}\implies X\geq \frac{2b^2}{ac}+\frac{ac}{b^2}$

Note that RHS of the last inequality will be the tightest lower bound when it is minimized. Minimizing it using AM-GM inequality, we have,

$\frac{2b^2}{ac}+\frac{ac}{b^2}\geq 2\sqrt{2}$

Comparing the bounds of the inequalities we obtained, we have,

$X\geq 2\sqrt{2}$

Hence follows the answer.

Just in case anyone wonders, equality occurs at $a=4^{1/4}=c~,~b=2^{1/4}$ for all the three inequalities (verification for equality case in AM-GM) and as such, the value we found in the end is indeed the minimum since the equality cases are matched.

Prasun Biswas - 6 years, 3 months ago

Yes, the condition was not necessary. Note that the inequality we are trying to prove is homogenous.

Daniel Liu - 6 years, 3 months ago

Yes, I see that. But I don't get what you're trying to imply by that. Does homogeneity have any special kind of impact on inequalities?

Prasun Biswas - 6 years, 3 months ago

@Prasun Biswas – Because the inequality is homogenous (both sides have degree 0) we can let $a'=ka$ , $b'=kb$ , and $c'=kc$ for any non-zero real number $k$ and yield an identical inequality in $a',b',c'$ . However, the condition itself, because it is non-homogeneous, turns into $\dfrac{1}{a^4}+\dfrac{1}{b^4}+\dfrac{1}{c^4}=\dfrac{1}{k^4}$

But $\dfrac{1}{k^4}$ can equal any non-zero real, so we essentially just got rid of the condition.

Daniel Liu - 6 years, 2 months ago

@Daniel Liu – Ah, I get it now. Thanks! :D

Prasun Biswas - 6 years, 2 months ago

I think the conditions are necessary. Because refer to my solution, if I had $(X)=2$ then the minimum would double

Trevor Arashiro - 6 years, 3 months ago

Check again! It remains the same. When the given condition is $2$ , you are also minimizing the double of what you're minimizing now.

Prasun Biswas - 6 years, 3 months ago

@Prasun Biswas – Oh, that's what you meant. I was talking about it changing the answer, not remaining as it is.

Trevor Arashiro - 6 years, 3 months ago

Same method

Figel Ilham - 6 years, 3 months ago

Nice solution! But all inequalities are not proven without the equality case :) although it is quite clear that there must be an equality case.

Joel Tan - 6 years, 3 months ago

A general result for the problem:

If $\dfrac{1}{a^4}+\dfrac{1}{b^4}+\dfrac{1}{c^4}=k, k\in\mathbb{R^+}$ , then equality holds iff,

$a=c=\alpha~,~b=2^{-1/4}\alpha~,~\alpha=\frac{\sqrt{2}}{k^{1/4}}$

Prasun Biswas - 6 years, 3 months ago

Nicely generalsied

Aakash Khandelwal - 5 years, 7 months ago

And Nice problem.

Is this original?

Dev Sharma - 5 years, 5 months ago

Kelvin Hong
Sep 6, 2018

This problem had been discussed long time ago, but I didn't find solution similar to mine, so below is my solution:

$\begin{aligned}\dfrac{a^4b^4+b^4c^4+c^4a^4}{a^3b^2c^3}&=\dfrac{ab^2}{c^3}+\dfrac{b^2c}{a^3}+\dfrac12\dfrac{ac}{b^2}+\dfrac12\dfrac{ac}{b^2}\\&\geq 4\sqrt[4]{\dfrac{ab^2}{c^3}\cdot\dfrac{b^2c}{a^3}\cdot\dfrac12\dfrac{ac}{b^2}\cdot\dfrac12\dfrac{ac}{b^2}}\\&=2\sqrt2\end{aligned}$

I'm surprised that I didn't use the constraint stated in the problem, but I found that the constraint is just narrow down the solution range, this minimum can achieve when $a=c=\sqrt2, b=\sqrt[4]2$ otherwise we just need $a=c=\sqrt[4]2b$ so I think there is no need to provide the constraint.

Emmanuel Torres
Feb 9, 2017

I don't know why. but I assumed a^4 = c^4 = 4, and b^4 = 2. And solved from there. There was symmetry between a and c, but not b because of b^2 in the denominator. 32/(8*√2) * 100 = 282.

Aakash Khandelwal
Oct 20, 2015

We need to find minimum value of ab^2c. Now use GM>=HM taking numbers as a^4,2 times 2*b^4 and c^4

Why isn't it B cubed?

The answer is 282.

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