Difference between revisions of "2001 IMO Shortlist Problems/A6"
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Therefore, <cmath>\Bigg(\Big(\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}}\Big)\Bigg)^2 \ge \frac{(a+b+c)^3}{a^3+b^3+c^3+24abc} \ge 1</cmath> | Therefore, <cmath>\Bigg(\Big(\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}}\Big)\Bigg)^2 \ge \frac{(a+b+c)^3}{a^3+b^3+c^3+24abc} \ge 1</cmath> | ||
− | Thus, <cmath> | + | Thus, <cmath>\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}} \ge 1</cmath> |
+ | |||
+ | -- Haozhe Yang | ||
== Resources == | == Resources == |
Revision as of 00:07, 28 March 2021
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
Alternative Solution
By Carlson's Inequality, we can know that
Then,
On the other hand, and
Then,
Therefore,
Thus,
-- Haozhe Yang