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

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

@@ Line 1: / Line 1: @@
-==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
 <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>.
-===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>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}}
-[[Category:Olympiad Number Theory Problems]]
+[[Category:Olympiad Algebra Problems]]
+[[Category:Olympiad Inequality Problems]]

2001 IMO (Problems) • Resources
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