Isn't this USAMO 2017 Problem 6

Algebra Level 4

Let $S$ be the maximum possible value of $\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}+\frac{d}{a^3+4}$ given that $a$ , $b$ , $c$ , $d$ are nonnegative real numbers such that $a+b+c+d=4$ . Given that $S$ can be written in the form $\frac{m}{n}$ where $m,n$ are coprime positive integers, find $100m+n$ .

The answer is 101.

1 solution

Sharky Kesa
Aug 20, 2017

Note that $\dfrac{1}{b^3+4} \leq \dfrac{1}{4}$ , so $\dfrac{a}{b^3+4} \leq \dfrac{a}{4}$ . Summing cyclically gives us $\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}+\frac{d}{a^3+4} \leq \dfrac{a}{4}+\dfrac{b}{4}+\dfrac{c}{4}+\dfrac{d}{4}=1$

Equality occurs at $(a,0,4-a,0)$ and its cyclic variants. Thus, $m=n=1$ , so $100m+n=101$ .

$a$ , $b$ , $c$ and $d$ are non-negative. Equality occurs when $(0,0,0,4)$ .

Chew-Seong Cheong - 3 years, 9 months ago

What's wrong with what I said?

Sharky Kesa - 3 years, 9 months ago

The problem says "given that $a$ , $b$ , $c$ , $d$ are nonnegative real numbers" but your solution: "Equality occurs at $(a,0,4-a,0)$ ".

Chew-Seong Cheong - 3 years, 9 months ago

@Chew-Seong Cheong – Those are non-negative reals though...

Sharky Kesa - 3 years, 9 months ago

@Sharky Kesa – Sorry, I am lost.

Chew-Seong Cheong - 3 years, 9 months ago

@Chew-Seong Cheong – My solution states all solutions of the form $(a, 0, 4-a, 0)$ and $(0, b, 0, 4-b)$ satisfy, including your solution, which is $b=0$ .

Sharky Kesa - 3 years, 9 months ago

@Sharky Kesa – Oh, I didn't see the $4-a$ . I thought it was $-a$ .

Chew-Seong Cheong - 3 years, 9 months ago

@Chew-Seong Cheong – Read "... and its cyclic variants"

Pi Han Goh - 3 years, 9 months ago

Isn't this USAMO 2017 Problem 6

The answer is 101.

1 solution

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