# Difference between revisions of "2012 JBMO Problems/Problem 1"

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The inequality now simplifies to <cmath>A^2+B^2+C^2+12 \geq 4(A+B+C)</cmath> | The inequality now simplifies to <cmath>A^2+B^2+C^2+12 \geq 4(A+B+C)</cmath> | ||

Note that because <math>a</math>, <math>b</math>, and <math>c</math> are positive real numbers less than <math>1</math>, <math>A</math>, <math>B</math>, and <math>C</math> are always positive real numbers. | Note that because <math>a</math>, <math>b</math>, and <math>c</math> are positive real numbers less than <math>1</math>, <math>A</math>, <math>B</math>, and <math>C</math> are always positive real numbers. | ||

− | Rearranging terms shows that this further simplifies to <cmath>(A^2-4A+4)+(B^2- | + | Rearranging terms shows that this further simplifies to <cmath>(A^2-4A+4)+(B^2-4B+4)+(C^2-4C+4)\geq0</cmath> which equals <cmath>(A-2)^2+(B-2)^2+(C-2)^2\geq0</cmath> |

By the [[trivial inequality]] we know that this is always true. Finally, we have equality when <cmath>A=B=C=2</cmath> and <cmath>\frac{2-2a}{a}=\frac{2-2b}{b}=\frac{2-2c}{c}=4</cmath> | By the [[trivial inequality]] we know that this is always true. Finally, we have equality when <cmath>A=B=C=2</cmath> and <cmath>\frac{2-2a}{a}=\frac{2-2b}{b}=\frac{2-2c}{c}=4</cmath> | ||

Solving the equations yields that equality holds when <math>\boxed{a=b=c=\frac{1}{3}}</math> | Solving the equations yields that equality holds when <math>\boxed{a=b=c=\frac{1}{3}}</math> | ||

Solution by Someonenumber011 :) | Solution by Someonenumber011 :) |

## Revision as of 18:30, 22 December 2020

## Section 1

Let be positive real numbers such that . Prove that When does equality hold?

## Solution

The LHS rearranges to . Since we have that . Therefore, the LHS rearranges again to .

Now, distribute the on the RHS into the parenthesis and multiply the LHS and RHS by 2 to get Let and similarly for and .

The inequality now simplifies to Note that because , , and are positive real numbers less than , , , and are always positive real numbers. Rearranging terms shows that this further simplifies to which equals By the trivial inequality we know that this is always true. Finally, we have equality when and Solving the equations yields that equality holds when

Solution by Someonenumber011 :)