Difference between revisions of "2001 IMO Problems/Problem 2"
(→2 Alternate Solutions using Jensen's) |
Jerry yang (talk | contribs) (→Solution) |
||
(2 intermediate revisions by 2 users not shown) | |||
Line 17: | Line 17: | ||
− | === Alternate Solution using | + | === Alternate Solution using Hölder's === |
− | By | + | By Hölder's inequality, |
<math>\left(\sum_{cyc}\frac{a}{\sqrt{a^{2}+8bc}}\right)\left(\sum_{cyc}\frac{a}{\sqrt{a^{2}+8bc}}\right)\left(\sum_{cyc} a(a^{2}+8bc)\right)\ge (a+b+c)^{3}</math> | <math>\left(\sum_{cyc}\frac{a}{\sqrt{a^{2}+8bc}}\right)\left(\sum_{cyc}\frac{a}{\sqrt{a^{2}+8bc}}\right)\left(\sum_{cyc} a(a^{2}+8bc)\right)\ge (a+b+c)^{3}</math> | ||
Thus we need only show that | Thus we need only show that | ||
Line 30: | Line 30: | ||
<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. | <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 Jensen's === | + | === Alternate Solution 2 using Jensen's === |
We can rewrite | We can rewrite | ||
<cmath>\frac{a}{\sqrt{a^2+8bc}}+\frac{b}{\sqrt{b^2+8ac}}+\frac{c}{\sqrt{c^2+8ab}}</cmath> | <cmath>\frac{a}{\sqrt{a^2+8bc}}+\frac{b}{\sqrt{b^2+8ac}}+\frac{c}{\sqrt{c^2+8ab}}</cmath> | ||
Line 71: | Line 71: | ||
But that is true by AM-GM or Muirhead. Thus, proved. <math>\Box</math> | But that is true by AM-GM or Muirhead. Thus, proved. <math>\Box</math> | ||
+ | |||
+ | === Alternate Solution using Carlson === | ||
+ | |||
+ | By Carlson's Inequality, we can know that <cmath>\Bigg(\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}}\Bigg)\Bigg(\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}}\Bigg)\Big((a^3+8abc)+(b^3+8abc)+(c^3+8abc)\Big) \ge (a+b+c)^3</cmath> | ||
+ | |||
+ | Then, <cmath>\Bigg(\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}}\Bigg)^2 \ge \frac{(a+b+c)^3}{a^3+b^3+c^3+24abc}</cmath> | ||
+ | |||
+ | On the other hand, <cmath>3a^2b+3b^2c+3c^2a \ge 9abc</cmath> and <cmath>3ab^2+3bc^2+3ca^2 \ge 9abc</cmath> | ||
+ | |||
+ | Then, <cmath>(a+b+c)^3 = a^3+b^3+c^3+3(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2)+6abc \ge a^3+b^3+c^3+24abc</cmath> | ||
+ | |||
+ | Therefore, <cmath>\Bigg(\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}}\Bigg)^2 \ge \frac{(a+b+c)^3}{a^3+b^3+c^3+24abc} \ge 1</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 | ||
== See also == | == See also == |
Revision as of 00:16, 28 March 2021
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, : Chebycheff'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 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 |