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

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== Solution ==
 
== Solution ==
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{{solution}}
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We will use the Jenson's inequality.
  
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Now, normalize the inequality by assuming <math>a+b+c=1</math>
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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>.
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Thus, we have
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<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>
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Thus, we only need to show that <math>a^3+b^3+c^3+24abc \leq 1</math> i.e.
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<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>
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<cmath>i.e.  (a+b+c)(ab+bc+ca) \geq 9abc</cmath>
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Which is true since
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<cmath>(a+b+c)(ab+bc+ca) \geq (3\sqrt[3]{abc})(3\sqrt[3]{a^{2}b^{2}c^{2}}) = 9abc</cmath>
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The last part follows by the AM-GM inequality.
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Equality holds if <math>a=b=c</math>
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== Alternative Solution ==
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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>
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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>
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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>
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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>
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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>
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Thus, <cmath>\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}} \ge 1</cmath>
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-- Haozhe Yang
  
 
== Resources ==
 
== Resources ==

Latest revision as of 23:12, 27 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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