Difference between revisions of "2001 IMO Problems/Problem 2"
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<cmath>\sum_{cyc} \frac{a}{\sqrt{a^{2}+8bc}} \ge \sum_{cyc} \frac{a}{\sqrt{S+6bc}}</cmath> | <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 ^{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^{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>\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> | <cmath>=(a+b+c)\sqrt{\frac{3}{3(a+b+c)^{2}}}=1</cmath> | ||
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Equality holds if <math>a=b=c=1</math>. | 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>: | + | 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 | + | === 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 | ||
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which is the same as | 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> | <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 + | + | 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> | <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. | ||
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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 == |
Latest revision as of 10:04, 21 August 2024
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 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 |