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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 ==
−
{{solution}}
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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]]
−
 
−
The leader of the Bulgarian team had come up with a generalization of this with putting k in place of 8 and replacing 1 in the RHS by 3/sq root(1+k)
 

Latest revision as of 00:12, 28 March 2021

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

\[(a+b+c)(ab+bc+ca) \geq (3\sqrt[3]{abc})(3\sqrt[3]{a^{2}b^{2}c^{2}}) = 9abc\] The last part follows by the AM-GM inequality.

Equality holds if $a=b=c$

Alternative Solution

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

Resources

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