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Difference between revisions of "2001 IMO Problems/Problem 2"

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<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.
  
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=== Alternate Solution using Jensen's ===
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=== 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>  

Revision as of 14:38, 29 January 2019

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 $a=b=c=1$.

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


Alternate Solution using Holder's

By Holder'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 $(a+b)(b+c)(c+a)\ge 8abc$.

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 $f(x)=\sqrt{\frac{x^3}{x^3 + abc}}$. Then f is concave, 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 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 \[a^{\frac{14}{3}}+a^{2}b^{\frac{8}{3}}+a^{2}c^{\frac{8}{3}}+2a^{\frac{11}{3}}b^{\frac{4}{3}}+2a^{\frac{11}{3}}c^{\frac{4}{3}}+2a^{2}b^{\frac{4}{3}}c^{\frac{4}{3}} \geq a^{\frac{14}{3}}+8a^{\frac{8}{3}}bc\] After cancelling the $a^{\frac{14}{3}}$ term, we apply AM-GM to RHS and obtain \[a^{2}b^{\frac{8}{3}}+a^{2}c^{\frac{8}{3}}+2a^{\frac{11}{3}}b^{\frac{4}{3}}+2a^{\frac{11}{3}}c^{\frac{4}{3}}+2a^{2}b^{\frac{4}{3}}c^{\frac{4}{3}} \geq 8(a^{\frac{64}{3}}b^8c^8)^{\frac{1}{8}}=8a^{\frac{8}{3}}bc\] 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. $\Box$

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
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