Difference between revisions of "2001 IMO Shortlist Problems/A6"

Revision as of 12:37, 3 September 2017

Problem

Prove that for all positive real numbers $a,b,c$ ,

$\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}} \geq 1.$

Generalization

The leader of the Bulgarian team had come up with the following generalization to the inequality:

$\frac {a}{\sqrt {a^2 + kbc}} + \frac {b}{\sqrt {b^2 + kca}} + \frac {c}{\sqrt {c^2 + kab}} \geq \frac{3}{\sqrt{1+k}}.$

Solution

We will use the Jenson's inequality.

Now, normalize the inequality by assuming $a+b+c=1$

Consider the function $f(x)=\frac{1}{\sqrt{x}}$ . Note that this function is convex and monotonically decreasing which implies that if $a > b$ , then $f(a) < f(b)$ .

Thus, we have

$\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}} = af(a^2+8bc)+bf(b^2+8ca)+cf(c^2+8ab) \geq f(a^3+b^3+c^3+24abc)$

Thus, we only need to show that $a^3+b^3+c^3+24abc \leq 1$ i.e.

$a^3+b^3+c^3+24abc \leq (a+b+c)^3=a^3+b^3+c^3+3(a+b+c)(ab+bc+ca)-3abc$

$i.e. (a+b+c)(ab+bc+ca) \geq 9abc$

Which is true since

$(a+b+c)(ab+bc+ca)=(a+b)(b+c)(c+a)+abc \geq 2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ca}+abc=9abc$ The last part follows by the AM-GM inequality.

Equality holds if $a=b=c$

Resources

@@ Line 23: / Line 23: @@
 <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.   (ab+bc+ca) \geq 9abc</cmath>
+<cmath>i.e.   (a+b+c)(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>
+<cmath>(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.

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