Difference between revisions of "1995 IMO Problems/Problem 2"
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=== Solution 4 === | === Solution 4 === | ||
+ | After the setting <math>a=\frac{1}{x}, b=\frac{1}{y}, c=\frac{1}{z},</math> and as <math>abc=1</math> so <math>\left(\frac{1}{a}\cdot\frac{1}{b}\cdot\frac{1} {c}=1\right)</math> concluding <math>x y z=1 .</math> | ||
+ | |||
+ | <math>\textsf{Claim}:</math> | ||
+ | <cmath> | ||
+ | \frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \geq \frac{3}{2} | ||
+ | </cmath> | ||
+ | By Titu Lemma, | ||
+ | <cmath> | ||
+ | \implies\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \geq \frac{(x+y+z)^{2}}{2(x+y+z)} | ||
+ | </cmath> | ||
+ | <cmath> | ||
+ | \implies\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \geq \frac{(x+y+z)}{2} | ||
+ | </cmath> | ||
+ | Now by AM-GM we know that <cmath> (x+y+z)\geq3\sqrt[3]{xyz} | ||
+ | </cmath>and <math>xyz=1</math> which concludes to <math>\implies (x+y+z)\geq3\sqrt[3]{1}</math> | ||
+ | |||
+ | Therefore we get | ||
+ | |||
+ | <cmath> | ||
+ | \implies\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \geq \frac{3}{2} | ||
+ | </cmath>Hence our claim is proved ~~ Aritra12 | ||
+ | |||
+ | === Solution 5 === | ||
Proceed as in Solution 1, to arrive at the equivalent inequality | Proceed as in Solution 1, to arrive at the equivalent inequality | ||
<cmath> \frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} \ge \frac{3}{2} . </cmath> | <cmath> \frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} \ge \frac{3}{2} . </cmath> | ||
− | But we know that <cmath>x + y + z \ge 3xyz | + | But we know that <cmath>x + y + z \ge 3xyz = 3</cmath> by AM-GM. Furthermore, |
<cmath> (x + y + y + z + x + z) (\frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y}) \ge (x + y + z)^2 </cmath> | <cmath> (x + y + y + z + x + z) (\frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y}) \ge (x + y + z)^2 </cmath> | ||
by Cauchy-Schwarz, and so dividing by <math>2(x + y + z)</math> gives | by Cauchy-Schwarz, and so dividing by <math>2(x + y + z)</math> gives | ||
− | <cmath> | + | <cmath> |
as desired. | as desired. | ||
− | === Solution | + | === Solution 6 === |
Without clever substitutions, and only AM-GM! | Without clever substitutions, and only AM-GM! | ||
− | Note that <math>abc = 1 \implies a = \frac{1}{bc}</math>. The cyclic sum becomes <math>\sum_{cyc}\frac{(bc)^3}{b + c}</math>. Note that by AM-GM, the cyclic sum is greater than or equal to <math>3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^{\frac13}</math>. We now see that we have the three so we must be on the right path. We now only need to show that <math>\frac32 \geq 3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^\frac13</math>. Notice that by AM-GM, <math>a + b \geq 2\sqrt{ab}</math>, <math>b + c \geq 2\sqrt{bc}</math>, and <math>a + c \geq 2\sqrt{ac}</math>. Thus, we see that <math>(a+b)(b+c)(a+c) \geq 8</math>, concluding that <math>\sum_{cyc} \frac{(bc)^3}{b+c} \geq \frac32 \geq 3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^{\frac13}</math> | + | Note that <math>abc = 1 \implies a = \frac{1}{bc}</math>. The cyclic sum becomes <math>\sum_{cyc}\frac{(bc)^3}{b + c}</math>. Note that by AM-GM, the cyclic sum is greater than or equal to <math>3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^{\frac13}</math>. We now see that we have the three so we must be on the right path. We now only need to show that <math>\frac32 \geq 3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^\frac13</math>. Notice that by AM-GM, <math>a + b \geq 2\sqrt{ab}</math>, <math>b + c \geq 2\sqrt{bc}</math>, and <math>a + c \geq 2\sqrt{ac}</math>. Thus, we see that <math>(a+b)(b+c)(a+c) \geq 8</math>, concluding that <math>\sum_{cyc} \frac{(bc)^3}{b+c} \geq \frac32 \geq 3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^{\frac13}</math>. |
+ | |||
+ | ^ This solution is incorrect, as it does not prove inequalities in the right direction. Proving that <math>A \geq B</math>, and <math>C \geq B</math> does not show that <math>A \geq C \geq B</math>. | ||
+ | |||
+ | === Solution 7 from Brilliant Wiki (Muirheads) ==== | ||
+ | https://brilliant.org/wiki/muirhead-inequality/ | ||
+ | |||
+ | |||
+ | === Solution 8 (fast Titu's Lemma no substitutions) === | ||
+ | Rewrite <math>\frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} + \frac{1}{c^3(a+b)}</math> as <math>\frac{(1/a)^2}{a(b+c)} + \frac{(1/b)^2}{b(a+c)} + \frac{(1/c)^2}{c(a+b)}</math>. | ||
+ | |||
+ | Now applying Titu's lemma yields <math>\frac{(1/a)^2}{a(b+c)} + \frac{(1/b)^2}{b(a+c)} + \frac{(1/c)^2}{c(a+b)} \geq \frac{(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})^2}{a(b+c) + b(a+c) + c(a+b)} = \frac{(ab + bc + ca)^2}{2(ab + bc + ca)} = \frac{ab + bc + ca}{2}</math>. | ||
+ | |||
+ | Now applying the AM-GM inequality on <math>ab + bc +ca \geq 3((abc)^2)^{\frac{1}{3}} = 3</math>. The result now follows. | ||
+ | |||
+ | Note: <math>ab + bc + ca = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}</math>, because <math>abc = 1</math>. (Why? Because <math>a = \frac{1}{bc}</math>, and hence <math>\frac{1}{a} = bc</math>). | ||
+ | |||
+ | ~th1nq3r | ||
+ | |||
+ | Scroll all the way down | ||
{{alternate solutions}} | {{alternate solutions}} | ||
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[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] | ||
+ | {{IMO box|year=1995|num-b=1|num-a=3}} |
Latest revision as of 00:01, 26 July 2024
Contents
[hide]Problem
(Nazar Agakhanov, Russia) Let be positive real numbers such that . Prove that
Solution
Solution 1
We make the substitution , , . Then Since and are similarly sorted sequences, it follows from the Rearrangement Inequality that By the Power Mean Inequality, Symmetric application of this argument yields Finally, AM-GM gives us as desired.
Solution 2
We make the same substitution as in the first solution. We note that in general, It follows that and are similarly sorted sequences. Then by Chebyshev's Inequality, By AM-GM, , and by Nesbitt's Inequality, The desired conclusion follows.
Solution 3
Without clever substitutions: By Cauchy-Schwarz, Dividing by gives by AM-GM.
Solution 3b
Without clever notation: By Cauchy-Schwarz,
Dividing by and noting that by AM-GM gives as desired.
Solution 4
After the setting and as so concluding
By Titu Lemma, Now by AM-GM we know that and which concludes to
Therefore we get
Hence our claim is proved ~~ Aritra12
Solution 5
Proceed as in Solution 1, to arrive at the equivalent inequality But we know that by AM-GM. Furthermore, by Cauchy-Schwarz, and so dividing by gives as desired.
Solution 6
Without clever substitutions, and only AM-GM!
Note that . The cyclic sum becomes . Note that by AM-GM, the cyclic sum is greater than or equal to . We now see that we have the three so we must be on the right path. We now only need to show that . Notice that by AM-GM, , , and . Thus, we see that , concluding that .
^ This solution is incorrect, as it does not prove inequalities in the right direction. Proving that , and does not show that .
Solution 7 from Brilliant Wiki (Muirheads) =
https://brilliant.org/wiki/muirhead-inequality/
Solution 8 (fast Titu's Lemma no substitutions)
Rewrite as .
Now applying Titu's lemma yields .
Now applying the AM-GM inequality on . The result now follows.
Note: , because . (Why? Because , and hence ).
~th1nq3r
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Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
Resources
1995 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |