Difference between revisions of "2001 IMO Problems/Problem 2"
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== Solution == | == Solution == | ||
− | === Solution using | + | Firstly, <math>a^{2}+8bc=(a^{2}+2bc)+6bc \leq^{AG} (a^{2}+b^{2}+c^{2})+6bc=S+6bc</math> (where <math>S=a^{2}+b^{2}+c^{2}</math>) and its cyclic variations. |
− | By | + | Next note that <math>(a,b,c)</math> and <math>\left( \frac{1}{\sqrt{S+6bc}}, \frac{1}{\sqrt{S+6ca}}, \frac{1}{\sqrt{S+6ab}} \right)</math> are similarly oriented sequences. Thus |
− | <math>\left(\ | + | <cmath>\sum_{cyc} \frac{a}{\sqrt{a^{2}+8bc}} \ge \sum_{cyc} \frac{a}{\sqrt{S+6bc}}</cmath> |
+ | <cmath>\geq ^{Cheff} \frac{1}{3}(a+b+c)\left( \frac{1}{\sqrt{S+6bc}}+\frac{1}{\sqrt{S+6ca}}+\frac{1}{\sqrt{S+6ab}} \right)</cmath> | ||
+ | <cmath>\geq^{AH}\frac{1}{3}(a+b+c) \left( \frac{9}{\sqrt{S+6bc}+\sqrt{S+6ca}+\sqrt{S+6ab}} \right)</cmath> | ||
+ | <cmath>\geq^{QA} (a+b+c) \sqrt{\frac{3}{(S+6bc)+(S+6ca)+(S+6ab)}}</cmath> | ||
+ | <cmath>=(a+b+c)\sqrt{\frac{3}{3(a+b+c)^{2}}}=1</cmath> | ||
+ | Hence the inequality has been established. | ||
+ | Equality holds if <math>a=b=c</math>. | ||
+ | |||
+ | Notation: <math>AG</math>: AM-GM inequality, <math>AH</math>: AM-HM inequality, <math>Cheff</math>: Chebyshev's inequality, <math>QA</math>: QM-AM inequality / RMS inequality | ||
+ | |||
+ | |||
+ | === Alternate Solution using Hölder's === | ||
+ | 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> | ||
Thus we need only show that | Thus we need only show that | ||
<math>(a+b+c)^{3}\ge a^{3}+b^{3}+c^{3}+24abc</math> | <math>(a+b+c)^{3}\ge a^{3}+b^{3}+c^{3}+24abc</math> | ||
Which is obviously true since <math>(a+b)(b+c)(c+a)\ge 8abc</math>. | Which is obviously true since <math>(a+b)(b+c)(c+a)\ge 8abc</math>. | ||
+ | === The Hölder's video solution === | ||
+ | https://youtu.be/jmXSmmfO7pQ?si=dxJ6At7KHlcn2NT5 [Video Solution by little fermat] | ||
=== Alternate Solution using Jensen's === | === Alternate Solution using Jensen's === | ||
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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 2 using Jensen's === | ||
+ | We can rewrite | ||
+ | <cmath>\frac{a}{\sqrt{a^2+8bc}}+\frac{b}{\sqrt{b^2+8ac}}+\frac{c}{\sqrt{c^2+8ab}}</cmath> | ||
+ | as | ||
+ | <cmath>\frac{a}{\sqrt{a^2+\frac{8abc}{a}}}+\frac{b}{\sqrt{b^2+\frac{8abc}{b}}}+\frac{c}{\sqrt{c^2+\frac{8abc}{c}}}</cmath> | ||
+ | which is the same as | ||
+ | <cmath>\frac{\sqrt{a^3}}{\sqrt{a^3+8abc}}+\frac{\sqrt{b^3}}{\sqrt{b^3+8abc}}+\frac{\sqrt{c^3}}{\sqrt{c^3+8abc}}</cmath> | ||
+ | Now let <math>f(x)=\sqrt{\frac{x^3}{x^3 + 8abc}}</math>. Then f is convex 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> | ||
+ | |||
+ | === 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 === | ||
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<cmath>\frac{a}{\sqrt{a^2+8bc}} \geq \frac{a^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}}</cmath> | <cmath>\frac{a}{\sqrt{a^2+8bc}} \geq \frac{a^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}}</cmath> | ||
Cross-multiplying, squaring both sides and expanding, we have | Cross-multiplying, squaring both sides and expanding, we have | ||
− | <cmath>a^{\frac{14}{3}}+a^{2}b^{\frac{8}{3}}+a^{2}c^{\frac{8}{3}}+2a^{\frac{ | + | <cmath>a^{\frac{14}{3}}+a^{2}b^{\frac{8}{3}}+a^{2}c^{\frac{8}{3}}+2a^{\frac{10}{3}}b^{\frac{4}{3}}+2a^{\frac{10}{3}}c^{\frac{4}{3}}+2a^{2}b^{\frac{4}{3}}c^{\frac{4}{3}} \geq a^{\frac{14}{3}}+8a^{\frac{8}{3}}bc</cmath> |
After cancelling the <math> a^{\frac{14}{3}}</math> term, we apply AM-GM to RHS and obtain | After cancelling the <math> a^{\frac{14}{3}}</math> term, we apply AM-GM to RHS and obtain | ||
− | <cmath>a^{2}b^{\frac{8}{3}}+a^{2}c^{\frac{8}{3}}+2a^{\frac{ | + | <cmath>a^{2}b^{\frac{8}{3}}+a^{2}c^{\frac{8}{3}}+2a^{\frac{10}{3}}b^{\frac{4}{3}}+2a^{\frac{10}{3}}c^{\frac{4}{3}}+2a^{2}b^{\frac{4}{3}}c^{\frac{4}{3}} \geq 8(a^{\frac{64}{3}}b^8c^8)^{\frac{1}{8}}=8a^{\frac{8}{3}}bc</cmath> |
as desired, completing the proof of the claim. | as desired, completing the proof of the claim. | ||
Similarly <math>\frac{b}{\sqrt{b^2+8ca}} \geq \frac{b^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}}</math> and <math>\frac{c}{\sqrt{c^2+8ab}} \geq \frac{c^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}}</math>. | Similarly <math>\frac{b}{\sqrt{b^2+8ca}} \geq \frac{b^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}}</math> and <math>\frac{c}{\sqrt{c^2+8ab}} \geq \frac{c^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}}</math>. | ||
Summing the three inequalities, we obtain the original inequality. | Summing the three inequalities, we obtain the original inequality. | ||
+ | |||
+ | === Alternate Solution using Cauchy === | ||
+ | |||
+ | We want to prove <cmath>\sum_{cyc}\dfrac{a}{\sqrt{a^2+8bc}}\ge 1</cmath> | ||
+ | |||
+ | Note that since this inequality is homogenous, assume <math>a+b+c=3</math>. | ||
+ | |||
+ | By Cauchy, <cmath>\left(\sum_{cyc}\dfrac{a}{\sqrt{a^2+8bc}}\right)\left(\sum_{cyc}a\sqrt{a^2+8bc}\right)\ge (a+b+c)^2=9</cmath> | ||
+ | |||
+ | Dividing both sides by <math>\sum_{cyc}a\sqrt{a^2+8bc}</math>, we see that we want to prove <cmath>\dfrac{9}{\sum\limits_{cyc}a\sqrt{a^2+8bc}}\ge 1</cmath> or equivalently <cmath>\sum\limits_{cyc}a\sqrt{a^2+8bc}\le 9</cmath> | ||
+ | |||
+ | Squaring both sides, we have <cmath>\left(\sum_{cyc}a\sqrt{a^2+8bc}\right)^2\le 81</cmath> | ||
+ | |||
+ | Now use Cauchy again to obtain <cmath>\left(\sum_{cyc}a\sqrt{a^2+8bc}\right)^2\le (a+b+c)\left(\sum_{cyc}a(a^2+8bc)\right)\le 81</cmath> | ||
+ | |||
+ | Since <math>a+b+c=3</math>, the inequality becomes <cmath>\sum_{cyc}a^3+8abc\le 27</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 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 == | ||
{{IMO box|year=2001|num-b=1|num-a=3}} | {{IMO box|year=2001|num-b=1|num-a=3}} | ||
+ | [[Category:Olympiad Algebra Problems]] | ||
[[Category:Olympiad Inequality Problems]] | [[Category:Olympiad Inequality Problems]] |
Latest revision as of 06:11, 22 October 2024
Problem
Let be positive real numbers. Prove that .
Contents
- 1 Problem
- 2 Solution
- 2.1 Alternate Solution using Hölder's
- 2.2 The Hölder's video solution
- 2.3 Alternate Solution using Jensen's
- 2.4 Alternate Solution 2 using Jensen's
- 2.5 Alternate Solution 3 using Jensen's
- 2.6 Alternate Solution using Isolated Fudging
- 2.7 Alternate Solution using Cauchy
- 2.8 Alternate Solution using Carlson
- 3 See also
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 .
The Hölder's video solution
https://youtu.be/jmXSmmfO7pQ?si=dxJ6At7KHlcn2NT5 [Video Solution by little fermat]
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 convex 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 |