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Difference between revisions of "2023 SSMO Team Round Problems/Problem 6"

(Created page with "==Problem== Suppose that <math>a,b,c</math> are positive reals satisfying<cmath>(a^3+4)(b^3+6)(c^3+8) = 8(a+b+c)^3.</cmath> Find the sum of all possible values of <math>\frac{...")
 
 
Line 3: Line 3:
  
 
==Solution==
 
==Solution==
 +
From Hölder's Inequality, we have:
 +
<cmath>(a^3 + 2 + 2)(2 + b^3 + 4)(4 + 4 + c^3) \geq 8(a + b + c)^3.</cmath>
 +
 +
Equality holds when:
 +
<cmath>a^3 : 2 : 2 = 2 : b^3 : 4 = 4 : 4 : c^3.</cmath>
 +
 +
From the first and third proportions, we get <math>a^3 = 2</math> and <math>c^3 = 2</math>. 
 +
From the first and second proportions, we get <math>a^3 = 1</math>.
 +
 +
This is a contradiction, so equality cannot occur. 
 +
Therefore, there are no real numbers <math>a, b, c</math> satisfying the equality condition, and the answer is:
 +
<cmath>\boxed{0}.</cmath>
 +
 +
~SMO_Team

Latest revision as of 21:19, 9 September 2025

Problem

Suppose that $a,b,c$ are positive reals satisfying\[(a^3+4)(b^3+6)(c^3+8) = 8(a+b+c)^3.\] Find the sum of all possible values of $\frac{bc}{a^2}.$ If you believe there are no solutions, put $0$ as your answer. If you believe the sum is infinity, put $1000$ as your answer.

Solution

From Hölder's Inequality, we have: \[(a^3 + 2 + 2)(2 + b^3 + 4)(4 + 4 + c^3) \geq 8(a + b + c)^3.\]

Equality holds when: \[a^3 : 2 : 2 = 2 : b^3 : 4 = 4 : 4 : c^3.\]

From the first and third proportions, we get $a^3 = 2$ and $c^3 = 2$. From the first and second proportions, we get $a^3 = 1$.

This is a contradiction, so equality cannot occur. Therefore, there are no real numbers $a, b, c$ satisfying the equality condition, and the answer is: \[\boxed{0}.\]

~SMO_Team

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