300+ followers problem

Algebra Level 5

$\large \dfrac{1}{x-y}+\dfrac{1}{y-z}+\dfrac{1}{x-z}$

Let $m$ be the minimum value of the above expression for reals $x > y > z$ given $(x - y)(y - z)(x - z) = 300$ .

Given that $m$ can be expressed as $\dfrac{1}{A}\sqrt[3]{\dfrac{B}{C}}$ where $A, B, C$ are positive integers and $\gcd(B,C)=1$ , with $B+C$ minimized.

Find $A+B+C$ .

The answer is 13.

1 solution

Tanishq Varshney
Jan 9, 2016

Quite frankly, I saw that and solved , @Khang Nguyen Thanh I have a doubt that why on earth is $4$ multiplied in the second step of the solution not 2 or 3 and why AM-GM is not applicable directly instead of multiplying with 4. Please reply soon.

@Tanishq Varshney : The equality holds iff $a=b$ or $\dfrac{a+b}{4ab}=\dfrac{1}{a+b}$ .

So AM-GM is not applicable directly instead of multiplying with $4$ .

Khang Nguyen Thanh - 5 years, 5 months ago

Oh sorry for troubling you, I got it, thanx!!

Tanishq Varshney - 5 years, 5 months ago

I solved it using AM GM but my ans is 110

Akarsh Kumar Srit - 5 years, 4 months ago

I took x-y = m , and y-z=n. Therefore rearranging the given expression we have ([a+b)^2 + 150/(a+b) + 150/(a+b)]/300. Now by applying Am>=Gm in numerator I got minimum value as 1/10 * (45/2)^1/3. Hence answer 57. Where am I wrong??:(

Aakash Khandelwal - 5 years, 3 months ago

@Aakash Khandelwal – I have rechecked my calculations, and the correct answer still be $\boxed{13}$ .

Let $x-y=a$ and $y-z=b$ . Thererefore, $x-z=a+b$ .

Note that $P=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{a+b}=4.\dfrac{a+b}{4ab}+\dfrac{1}{a+b}$ .

Applying AM-GM here on these five fractions, we get $P\ge5\sqrt[5]{\dfrac{(a+b)^3}{256a^4b^4}}$ .

Note that because $ab(a+b)=17$ , this is equivalent to $P\ge5\sqrt[5]{\dfrac{(a+b)^7}{256\times 300^4}}$ .

On the other hand, $300=ab(a+b)\le\dfrac{(a+b)^3}{4}$ or $a+b\ge\sqrt[3]{1200}$ .

This implies that $P\ge5\sqrt[5]{\dfrac{\sqrt[3]{1200^7}}{256\times 300^4}}=\dfrac{1}{2}\sqrt[3]{\dfrac{5}{6}}$

So $A=2; B=5; C=6$ and $A+B+C=\boxed{13}$ .

Khang Nguyen Thanh - 5 years, 3 months ago

I also did it using AM-GM...and my ans. Is also 110.

Rishabh Tiwari - 5 years, 3 months ago

I have rechecked my calculations, and the correct answer still be $\boxed{13}$ .