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 :)