Difference between revisions of "2001 IMO Problems/Problem 2"
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<cmath>\frac{a}{\sqrt{a^2+8bc}}+\frac{b}{\sqrt{b^2+8ac}}+\frac{c}{\sqrt{c^2+8ab}} \geq \frac{1}{\sqrt{a^3+b^3+c^3+24abc}}</cmath> | <cmath>\frac{a}{\sqrt{a^2+8bc}}+\frac{b}{\sqrt{b^2+8ac}}+\frac{c}{\sqrt{c^2+8ab}} \geq \frac{1}{\sqrt{a^3+b^3+c^3+24abc}}</cmath> | ||
but | but | ||
− | <cmath>1=(a+b+c)^3=a^3+b^3+c^3+6abc+3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b) \geq a^3+b^3+c^3+24abc</cmath> by | + | <cmath>1=(a+b+c)^3=a^3+b^3+c^3+6abc+3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b) \geq a^3+b^3+c^3+24abc</cmath> by AM-GM, and thus the inequality is proven. |
=== Alternate Solution using Isolated Fudging === | === Alternate Solution using Isolated Fudging === |
Revision as of 21:07, 16 April 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. 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 |