Difference between revisions of "2001 IMO Shortlist Problems/A6"
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Prove that for all positive real numbers <math>a,b,c</math>, | Prove that for all positive real numbers <math>a,b,c</math>, | ||
<center><math>\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}} \geq 1.</math></center> | <center><math>\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}} \geq 1.</math></center> | ||
+ | |||
+ | === Generalization === | ||
+ | The leader of the Bulgarian team had come up with the following generalization to the inequality: | ||
+ | |||
+ | <center><math>\frac {a}{\sqrt {a^2 + kbc}} + \frac {b}{\sqrt {b^2 + kca}} + \frac {c}{\sqrt {c^2 + kab}} \geq \frac{3}{\sqrt{1+k}}.</math></center> | ||
== Solution == | == Solution == | ||
− | {{ | + | We will use the Jenson's inequality. |
+ | |||
+ | Now, normalize the inequality by assuming <math>a+b+c=1</math> | ||
+ | |||
+ | Consider the function <math>f(x)=\frac{1}{\sqrt{x}}</math>. Note that this function is convex and monotonically decreasing which implies that if <math>a > b</math>, then <math>f(a) < f(b)</math>. | ||
+ | |||
+ | Thus, we have | ||
+ | |||
+ | <math>\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}} = af(a^2+8bc)+bf(b^2+8ca)+cf(c^2+8ab) \geq f(a^3+b^3+c^3+24abc)</math> | ||
+ | |||
+ | Thus, we only need to show that <math>a^3+b^3+c^3+24abc \leq 1</math> i.e. | ||
+ | |||
+ | <cmath>a^3+b^3+c^3+24abc \leq (a+b+c)^3=a^3+b^3+c^3+3(a+b+c)(ab+bc+ca)-3abc</cmath> | ||
+ | |||
+ | <cmath>i.e. (a+b+c)(ab+bc+ca) \geq 9abc</cmath> | ||
+ | |||
+ | Which is true since | ||
+ | |||
+ | <cmath>(a+b+c)(ab+bc+ca) \geq (3\sqrt[3]{abc})(3\sqrt[3]{a^{2}b^{2}c^{2}}) = 9abc</cmath> | ||
+ | The last part follows by the AM-GM inequality. | ||
+ | |||
+ | Equality holds if <math>a=b=c</math> | ||
+ | |||
+ | == Alternative Solution == | ||
+ | |||
+ | 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 | ||
== Resources == | == Resources == | ||
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[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] | ||
[[Category:Olympiad Inequality Problems]] | [[Category:Olympiad Inequality Problems]] | ||
− | |||
− |
Latest revision as of 00:12, 28 March 2021
Problem
Prove that for all positive real numbers ,
![$\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}} \geq 1.$](http://latex.artofproblemsolving.com/b/9/9/b99abc5062d845cddcb97bb99675fe9a4de3a2ee.png)
Generalization
The leader of the Bulgarian team had come up with the following generalization to the inequality:
![$\frac {a}{\sqrt {a^2 + kbc}} + \frac {b}{\sqrt {b^2 + kca}} + \frac {c}{\sqrt {c^2 + kab}} \geq \frac{3}{\sqrt{1+k}}.$](http://latex.artofproblemsolving.com/4/7/1/471af15648323ef1d44e3ee7252c1941f2ac8f40.png)
Solution
We will use the Jenson's inequality.
Now, normalize the inequality by assuming
Consider the function . Note that this function is convex and monotonically decreasing which implies that if
, then
.
Thus, we have
Thus, we only need to show that i.e.
Which is true since
The last part follows by the AM-GM inequality.
Equality holds if
Alternative Solution
By Carlson's Inequality, we can know that
Then,
On the other hand, and
Then,
Therefore,
Thus,
-- Haozhe Yang