Difference between revisions of "2001 IMO Shortlist Problems/A6"
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Thus, we only need to show that <math>a^3+b^3+c^3+24abc \leq 1</math> i.e. | Thus, we only need to show that <math>a^3+b^3+c^3+24abc \leq 1</math> i.e. | ||
− | <cmath>a^3+b^3+c^3+24abc \leq (a+b+c)^3=a^3+b^3+c^3+3(ab+bc+ca)-3abc</cmath> | + | <cmath>a^3+b^3+c^3+24abc \leq (a+b+c)^3=a^3+b^3+c^3+3(a+b+c)(ab+bc+ca)-3abc</cmath> |
<cmath>i.e. (ab+bc+ca) \geq 9abc</cmath> | <cmath>i.e. (ab+bc+ca) \geq 9abc</cmath> |
Revision as of 13:36, 3 September 2017
Contents
Problem
Prove that for all positive real numbers ,
![$\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}} \geq 1.$](http://latex.artofproblemsolving.com/b/9/9/b99abc5062d845cddcb97bb99675fe9a4de3a2ee.png)
Generalization
The leader of the Bulgarian team had come up with the following generalization to the inequality:
![$\frac {a}{\sqrt {a^2 + kbc}} + \frac {b}{\sqrt {b^2 + kca}} + \frac {c}{\sqrt {c^2 + kab}} \geq \frac{3}{\sqrt{1+k}}.$](http://latex.artofproblemsolving.com/4/7/1/471af15648323ef1d44e3ee7252c1941f2ac8f40.png)
Solution
We will use the Jenson's inequality.
Now, normalize the inequality by assuming
Consider the function . Note that this function is convex and monotonically decreasing which implies that if
, then
.
Thus, we have
Thus, we only need to show that i.e.
Which is true since
The last part follows by the AM-GM inequality.
Equality holds if