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

Revision as of 23:04, 27 December 2017

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 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. $\Box$

@@ Line 4: / Line 4: @@
 __TOC__
 == Solution ==
-=== Solution using Holder's ===
+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.
+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=1</math>.
+Notation: <math>AG</math>: AM-GM inequality, <math>AH</math>: AM-HM inequality, <math>Cheff</math>: Chebycheff's inequality, <math>QA</math>: QM-AM inequality / RMS inequality
+=== Alternate Solution using Holder's ===
 By Holder'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>

2001 IMO (Problems) • Resources
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