Difference between revisions of "2001 IMO Shortlist Problems/A6"
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== 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(ab+bc+ca)-3abc</cmath> | ||
+ | |||
+ | <cmath>i.e. (ab+bc+ca) \geq 9abc</cmath> | ||
+ | |||
+ | Which is true since | ||
+ | |||
+ | <cmath>(ab+bc+ca)=(a+b+c)(ab+bc+ca)=(a+b)(b+c)(c+a)+abc \geq 2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ca}+abc=9abc</cmath> | ||
+ | The last part follows by the AM-GM inequality. | ||
+ | |||
+ | Equality holds if <math>a=b=c</math> | ||
== Resources == | == Resources == |
Revision as of 11:01, 1 September 2017
Contents
Problem
Prove that for all positive real numbers ,
Generalization
The leader of the Bulgarian team had come up with the following generalization to the inequality:
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