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2001 IMO Problems/Problem 2: Difference between revisions

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==Problem==
== Problem ==
Let <math>a,b,c</math> be positive real numbers. Prove that
Let <math>a,b,c</math> be positive real numbers. Prove that <math>\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge 1</math>.
<math>\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge 1</math>
 
==Solution==
__TOC__
===Solution using Holder's===
== Solution ==
By Holder's inequality,
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.
<math>\left(\sum\frac{a}{\sqrt{a^{2}+8bc}}\right)\left(\sum\frac{a}{\sqrt{a^{2}+8bc}}\right)\left(\sum a(a^{2}+8bc)\right)\ge (a+b+c)^{3}</math>
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
<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>.
===Alternate Solution using Jensen's===
=== The Hölder's video solution ===
By Jensen's,  
https://youtu.be/jmXSmmfO7pQ?si=dxJ6At7KHlcn2NT5 [Video Solution by little fermat]
<math>\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge a^{3}+b^{3}+c^{3}+24abc</math>, so we need to prove <math>\frac{1}{\sqrt{a^{3}+b^{3}+c^{3}+24abc}}\ge 1</math>, which is obvious by [[RMS-AM-GM-HM]].
 
=== Alternate Solution using Jensen's ===
This inequality is homogeneous so we can assume without loss of generality <math>a+b+c=1</math> and apply Jensen's inequality for <math>f(x)=\frac{1}{\sqrt{x}}</math>, so we get:
<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
<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 ===
We claim that
<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
<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
<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.
 
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.
 
=== 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 ==
{{IMO box|year=2001|num-b=1|num-a=3}}
{{IMO box|year=2001|num-b=1|num-a=3}}
[[Category:Olympiad Number Theory Problems]]
 
[[Category:Olympiad Algebra Problems]]
[[Category:Olympiad Inequality Problems]]

Latest revision as of 05:11, 22 October 2024

Problem

Let $a,b,c$ be positive real numbers. Prove that $\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge 1$.

Solution

Firstly, $a^{2}+8bc=(a^{2}+2bc)+6bc \leq^{AG} (a^{2}+b^{2}+c^{2})+6bc=S+6bc$ (where $S=a^{2}+b^{2}+c^{2}$) and its cyclic variations. Next note that $(a,b,c)$ and $\left( \frac{1}{\sqrt{S+6bc}}, \frac{1}{\sqrt{S+6ca}}, \frac{1}{\sqrt{S+6ab}} \right)$ are similarly oriented sequences. Thus \[\sum_{cyc} \frac{a}{\sqrt{a^{2}+8bc}} \ge \sum_{cyc} \frac{a}{\sqrt{S+6bc}}\] \[\geq ^{Cheff} \frac{1}{3}(a+b+c)\left( \frac{1}{\sqrt{S+6bc}}+\frac{1}{\sqrt{S+6ca}}+\frac{1}{\sqrt{S+6ab}} \right)\] \[\geq^{AH}\frac{1}{3}(a+b+c) \left( \frac{9}{\sqrt{S+6bc}+\sqrt{S+6ca}+\sqrt{S+6ab}} \right)\] \[\geq^{QA} (a+b+c) \sqrt{\frac{3}{(S+6bc)+(S+6ca)+(S+6ab)}}\] \[=(a+b+c)\sqrt{\frac{3}{3(a+b+c)^{2}}}=1\] Hence the inequality has been established. Equality holds if $a=b=c$.

Notation: $AG$: AM-GM inequality, $AH$: AM-HM inequality, $Cheff$: Chebyshev's inequality, $QA$: QM-AM inequality / RMS inequality


Alternate Solution using Hölder's

By Hölder's inequality, $\left(\sum_{cyc}\frac{a}{\sqrt{a^{2}+8bc}}\right)\left(\sum_{cyc}\frac{a}{\sqrt{a^{2}+8bc}}\right)\left(\sum_{cyc} a(a^{2}+8bc)\right)\ge (a+b+c)^{3}$ Thus we need only show that $(a+b+c)^{3}\ge a^{3}+b^{3}+c^{3}+24abc$ Which is obviously true since $(a+b)(b+c)(c+a)\ge 8abc$.

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 $a+b+c=1$ and apply Jensen's inequality for $f(x)=\frac{1}{\sqrt{x}}$, so we get: \[\frac{a}{\sqrt{a^2+8bc}}+\frac{b}{\sqrt{b^2+8ac}}+\frac{c}{\sqrt{c^2+8ab}} \geq \frac{1}{\sqrt{a^3+b^3+c^3+24abc}}\] but \[1=(a+b+c)^3=a^3+b^3+c^3+6abc+3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b) \geq a^3+b^3+c^3+24abc\] by AM-GM, and thus the inequality is proven.

Alternate Solution 2 using Jensen's

We can rewrite \[\frac{a}{\sqrt{a^2+8bc}}+\frac{b}{\sqrt{b^2+8ac}}+\frac{c}{\sqrt{c^2+8ab}}\] as \[\frac{a}{\sqrt{a^2+\frac{8abc}{a}}}+\frac{b}{\sqrt{b^2+\frac{8abc}{b}}}+\frac{c}{\sqrt{c^2+\frac{8abc}{c}}}\] which is the same as \[\frac{\sqrt{a^3}}{\sqrt{a^3+8abc}}+\frac{\sqrt{b^3}}{\sqrt{b^3+8abc}}+\frac{\sqrt{c^3}}{\sqrt{c^3+8abc}}\] Now let $f(x)=\sqrt{\frac{x^3}{x^3 + 8abc}}$. Then f is convex and f is strictly increasing, so by Jensen's inequality and AM-GM, \[f(a) + f(b) + f(c) \geq 3f((\frac{1}{3})a + (\frac{1}{3})b + (\frac{1}{3})c)) \geq 3f(\sqrt[3]{abc}) = 3(\frac{1}{3}) =1\]

Alternate Solution 3 using Jensen's

Let $f : (0, \infty); f(x) = \frac{1}{x}$, $x_1 = \sqrt{a^{2} + 8bc}$, $x_2 = \sqrt{b^{2} + 8ac}$ and $x_3 = \sqrt{c^{2} + 8ab}$ f is convex so we can write: \[f(\frac{a}{x_1} + \frac{b}{x_2} + \frac{c}{x_3})\le af(x_1) + bf(x_2) + cf(x_3)\] let $\frac{a}{x_1} + \frac{b}{x_2} + \frac{c}{x_3} = t$, by substitustion: \[f(t)\le t\] \[\frac{1}{t}\le t\] we multiply both sides by t \[1\le t^{2}\] \[1\le t\] QED

Alternate Solution using Isolated Fudging

We claim that \[\frac{a}{\sqrt{a^2+8bc}} \geq \frac{a^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}}\] Cross-multiplying, squaring both sides and expanding, we have \[a^{\frac{14}{3}}+a^{2}b^{\frac{8}{3}}+a^{2}c^{\frac{8}{3}}+2a^{\frac{10}{3}}b^{\frac{4}{3}}+2a^{\frac{10}{3}}c^{\frac{4}{3}}+2a^{2}b^{\frac{4}{3}}c^{\frac{4}{3}} \geq a^{\frac{14}{3}}+8a^{\frac{8}{3}}bc\] After cancelling the $a^{\frac{14}{3}}$ term, we apply AM-GM to RHS and obtain \[a^{2}b^{\frac{8}{3}}+a^{2}c^{\frac{8}{3}}+2a^{\frac{10}{3}}b^{\frac{4}{3}}+2a^{\frac{10}{3}}c^{\frac{4}{3}}+2a^{2}b^{\frac{4}{3}}c^{\frac{4}{3}} \geq 8(a^{\frac{64}{3}}b^8c^8)^{\frac{1}{8}}=8a^{\frac{8}{3}}bc\] as desired, completing the proof of the claim.

Similarly $\frac{b}{\sqrt{b^2+8ca}} \geq \frac{b^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}}$ and $\frac{c}{\sqrt{c^2+8ab}} \geq \frac{c^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}}$. Summing the three inequalities, we obtain the original inequality.

Alternate Solution using Cauchy

We want to prove \[\sum_{cyc}\dfrac{a}{\sqrt{a^2+8bc}}\ge 1\]

Note that since this inequality is homogenous, assume $a+b+c=3$.

By Cauchy, \[\left(\sum_{cyc}\dfrac{a}{\sqrt{a^2+8bc}}\right)\left(\sum_{cyc}a\sqrt{a^2+8bc}\right)\ge (a+b+c)^2=9\]

Dividing both sides by $\sum_{cyc}a\sqrt{a^2+8bc}$, we see that we want to prove \[\dfrac{9}{\sum\limits_{cyc}a\sqrt{a^2+8bc}}\ge 1\] or equivalently \[\sum\limits_{cyc}a\sqrt{a^2+8bc}\le 9\]

Squaring both sides, we have \[\left(\sum_{cyc}a\sqrt{a^2+8bc}\right)^2\le 81\]

Now use Cauchy again to obtain \[\left(\sum_{cyc}a\sqrt{a^2+8bc}\right)^2\le (a+b+c)\left(\sum_{cyc}a(a^2+8bc)\right)\le 81\]

Since $a+b+c=3$, the inequality becomes \[\sum_{cyc}a^3+8abc\le 27\] after some simplifying.

But this equals \[(a+b+c)^3-3\left(\sum_{sym}a^2b\right)+18abc\le 27\] and since $a+b+c=3$ we just want to prove \[\left(\sum_{sym}a^2b\right)\ge 6abc\] after some simplifying.

But that is true by AM-GM or Muirhead. Thus, proved. $\Box$

Alternate Solution using Carlson

By Carlson's Inequality, we can know that \[\Bigg(\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}}\Bigg)\Bigg(\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}}\Bigg)\Big((a^3+8abc)+(b^3+8abc)+(c^3+8abc)\Big) \ge (a+b+c)^3\]

Then, \[\Bigg(\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}}\Bigg)^2 \ge \frac{(a+b+c)^3}{a^3+b^3+c^3+24abc}\]

On the other hand, \[3a^2b+3b^2c+3c^2a \ge 9abc\] and \[3ab^2+3bc^2+3ca^2 \ge 9abc\]

Then, \[(a+b+c)^3 = a^3+b^3+c^3+3(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2)+6abc \ge a^3+b^3+c^3+24abc\]

Therefore, \[\Bigg(\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}}\Bigg)^2 \ge \frac{(a+b+c)^3}{a^3+b^3+c^3+24abc} \ge 1\]

Thus, \[\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}} \ge 1\]

-- Haozhe Yang

See also

2001 IMO (Problems) • Resources
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Problem 1 		1 • 2 • 3 • 4 • 5 • 6 Followed by
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All IMO Problems and Solutions
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