Difference between revisions of "2001 IMO Problems/Problem 2"
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But this equals <cmath>(a+b+c)^3-3\left(\sum_{sym}a^2b\right)+18abc\le 27</cmath> and since <math>a+b+c=3</math> we just want to prove <cmath>\left(\sum_{sym}a^2b\right)\ge 6abc</cmath> after some simplifying. | But this equals <cmath>(a+b+c)^3-3\left(\sum_{sym}a^2b\right)+18abc\le 27</cmath> and since <math>a+b+c=3</math> we just want to prove <cmath>\left(\sum_{sym}a^2b\right)\ge 6abc</cmath> after some simplifying. | ||
− | But that is true by AM-GM. Thus, proved. <math>\Box</math> | + | But that is true by AM-GM or Muirhead. Thus, proved. <math>\Box</math> |
== See also == | == See also == |
Revision as of 16:00, 19 August 2015
Problem
Let be positive real numbers. Prove that .
Contents
Solution
Solution using Holder's
By Holder'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 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.
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 |