Difference between revisions of "2001 IMO Problems/Problem 2"
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Now let <math>f(x)=\sqrt{\frac{x^3}{x^3 + abc}}</math>. Then f is concave, and f is strictly increasing, so by Jensen's inequality and AM-GM, | Now let <math>f(x)=\sqrt{\frac{x^3}{x^3 + abc}}</math>. Then f is concave, and f is strictly increasing, so by Jensen's inequality and AM-GM, | ||
<cmath>f(a) + f(b) + f(c) \geq 3f((\frac{1}{3})a + (\frac{1}{3})b + (\frac{1}{3})c)) \geq 3f(\sqrt[3]{abc}) = 3(\frac{1}{3}) =1</cmath> | <cmath>f(a) + f(b) + f(c) \geq 3f((\frac{1}{3})a + (\frac{1}{3})b + (\frac{1}{3})c)) \geq 3f(\sqrt[3]{abc}) = 3(\frac{1}{3}) =1</cmath> | ||
+ | |||
+ | === Alternate Solution 3 using Jensen's === | ||
+ | Let <math>f : (0, \infty); f(x) = \frac{1}{x}</math>, <math>x_1 = \sqrt{a^{2} + 8bc}</math>, <math>x_2 = \sqrt{b^{2} + 8ac}</math> and <math>x_3 = \sqrt{c^{2} + 8ab}</math> | ||
+ | f is convex so we can write: | ||
+ | <cmath>f(\frac{a}{x_1} + \frac{b}{x_2} + \frac{c}{x_3})\le af(x_1) + bf(x_2) + cf(x_3)</cmath> | ||
+ | let <math>\frac{a}{x_1} + \frac{b}{x_2} + \frac{c}{x_3} = t</math>, by substitustion: | ||
+ | <cmath>f(t)\le t</cmath> | ||
+ | <cmath>\frac{1}{t}\le t</cmath> we multiply both sides by t | ||
+ | <cmath>1\le t^{2}</cmath> | ||
+ | <cmath>1\le t</cmath> QED | ||
=== Alternate Solution using Isolated Fudging === | === Alternate Solution using Isolated Fudging === |
Revision as of 15:08, 20 February 2023
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 |