Difference between revisions of "2001 IMO Problems/Problem 2"
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__TOC__ | __TOC__ | ||
== 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=1</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> | ||
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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 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 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 + 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> | ||
=== Alternate Solution using Isolated Fudging === | === Alternate Solution using Isolated Fudging === | ||
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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> |
+ | |||
+ | === 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 == |
Latest revision as of 04:59, 5 February 2022
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 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 |