2001 IMO Problems/Problem 2
Problem
Let be positive real numbers. Prove that .
Contents
Solution
Firstly, (where ) and its cyclic variations. Next note that and are similarly oriented sequences. Thus Hence the inequality has been established. Equality holds if .
Notation: : AM-GM inequality, : AM-HM inequality, : Chebyshev's inequality, : QM-AM inequality / RMS inequality
Alternate Solution using Hölder's
By Hölder's inequality, Thus we need only show that Which is obviously true since .
Alternate Solution using Jensen's
This inequality is homogeneous so we can assume without loss of generality and apply Jensen's inequality for , so we get: but by AM-GM, and thus the inequality is proven.
Alternate Solution 2 using Jensen's
We can rewrite as which is the same as Now let . Then f is concave, and f is strictly increasing, so by Jensen's inequality and AM-GM,
Alternate Solution 3 using Jensen's
Let , , and f is convex so we can write: let , by substitustion: we multiply both sides by t QED
Alternate Solution using Isolated Fudging
We claim that Cross-multiplying, squaring both sides and expanding, we have After cancelling the term, we apply AM-GM to RHS and obtain as desired, completing the proof of the claim.
Similarly and . Summing the three inequalities, we obtain the original inequality.
Alternate Solution using Cauchy
We want to prove
Note that since this inequality is homogenous, assume .
By Cauchy,
Dividing both sides by , we see that we want to prove or equivalently
Squaring both sides, we have
Now use Cauchy again to obtain
Since , the inequality becomes after some simplifying.
But this equals and since we just want to prove after some simplifying.
But that is true by AM-GM or Muirhead. Thus, proved.
Alternate Solution using Carlson
By Carlson's Inequality, we can know that
Then,
On the other hand, and
Then,
Therefore,
Thus,
-- Haozhe Yang
See also
2001 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |