Difference between revisions of "2012 JBMO Problems/Problem 1"
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The LHS rearranges to <math>\frac{b+c}{a} + \frac{a+c}{b} + \frac{a+b}{c} + 6</math>. Since <math>b+c=1-a</math> we have that <math>\frac{b+c}{a}=\frac{1-a}{a}</math>. Therefore, the LHS rearranges again to <math>\frac{1-a}{a}+\frac{1-b}{b}+\frac{1-c}{c}+6</math>. | The LHS rearranges to <math>\frac{b+c}{a} + \frac{a+c}{b} + \frac{a+b}{c} + 6</math>. Since <math>b+c=1-a</math> we have that <math>\frac{b+c}{a}=\frac{1-a}{a}</math>. Therefore, the LHS rearranges again to <math>\frac{1-a}{a}+\frac{1-b}{b}+\frac{1-c}{c}+6</math>. | ||
− | Now, | + | Now, distribute the <math>\sqrt{2}</math> on the RHS into the parenthesis and multiply the LHS and RHS by 2 to get <cmath>\frac{2-2a}{a} + \frac{2-2b}{b} + \frac{2-2c}{c}+12 \geq 4(\sqrt{\frac{2-2a}{a}} + \sqrt{\frac{2-2b}{b}} + \sqrt{\frac{2-2c}{c}})</cmath> Let <math>\sqrt{\frac{2-2a}{a}} = A</math> and similarly for <math>B</math> and <math>C</math>. |
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> |
Revision as of 18:23, 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
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 :)