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Difference between revisions of "2014 USAJMO Problems/Problem 1"

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==Problem==
 
==Problem==
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Let <math>a</math>, <math>b</math>, <math>c</math> be real numbers greater than or equal to <math>1</math>. Prove that <cmath>\min{\left (\frac{10a^2-5a+1}{b^2-5b+1},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right )}\leq abc </cmath>
+
Let <math>a</math>, <math>b</math>, <math>c</math> be real numbers greater than or equal to <math>1</math>. Prove that <cmath>\min{\left (\frac{10a^2-5a+1}{b^2-5b+10},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right )}\leq abc </cmath>
 +
 
 
==Solution==
 
==Solution==
−
Since <math>(a-1)^5\geqslant 0</math>,
+
Since <math>(a-1)^5\ge 0</math>,
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<cmath>a^5-5a^4+10a^3-10a^2+5a-1\geqslant 0</cmath>
+
<cmath>a^5-5a^4+10a^3-10a^2+5a-1\ge 0</cmath>
 
or
 
or
−
<cmath>10a^2-5a+1\leqslant a^3(a^2-5a+10)</cmath>
+
<cmath>10a^2-5a+1\le a^3(a^2-5a+10)</cmath>
−
Since <math>a^2-5a+10=\left( a-\dfrac{5}{2}\right)^2 +\dfrac{15}{4}\geqslant 0</math>,
+
Since <math>a^2-5a+10=\left( a-\dfrac{5}{2}\right)^2 +\dfrac{15}{4}>0</math>,
−
<cmath> \frac{10a^2-5a+1}{a^2-5a+10}\leqslant a^3 </cmath>
+
<cmath> \frac{10a^2-5a+1}{a^2-5a+10}\le a^3 </cmath>
−
Also note that <math>10a^2-5a+1=10\left( a-\dfrac{1}{4}\right)^2+\dfrac{3}{8}\geqslant 0</math>,
+
Also note that <math>10a^2-5a+1=10\left( a-\dfrac{1}{4}\right)^2+\dfrac{3}{8}> 0</math>,
 
We conclude
 
We conclude
−
<cmath>0\leqslant\frac{10a^2-5a+1}{a^2-5a+10}\leqslant a^3</cmath>
+
<cmath>0\le \frac{10a^2-5a+1}{a^2-5a+10}\le a^3</cmath>
 
Similarly,  
 
Similarly,  
−
<cmath>0\leqslant\frac{10b^2-5b+1}{b^2-5b+10}\leqslant b^3</cmath>
+
<cmath>0\le \frac{10b^2-5b+1}{b^2-5b+10}\le b^3</cmath>
−
<cmath>0\leqslant\frac{10c^2-5c+1}{c^2-5c+10}\leqslant c^3</cmath>
+
<cmath>0\le \frac{10c^2-5c+1}{c^2-5c+10}\le c^3</cmath>
−
So <cmath>\left(\frac{10a^2-5a+1}{a^2-5a+10}\right)\left(\frac{10b^2-5b+1}{b^2-5b+10}\right)\left(\frac{10c^2-5c+1}{c^2-5c+10}\right)\leqslant a^3b^3c^3</cmath>
+
So <cmath>\left(\frac{10a^2-5a+1}{a^2-5a+10}\right)\left(\frac{10b^2-5b+1}{b^2-5b+10}\right)\left(\frac{10c^2-5c+1}{c^2-5c+10}\right)\le a^3b^3c^3</cmath>
 
or
 
or
−
<cmath>\left(\frac{10a^2-5a+1}{b^2-5b+10}\right)\left(\frac{10b^2-5b+1}{c^2-5c+10}\right)\left(\frac{10c^2-5c+1}{a^2-5a+10}\right) \leqslant(abc)^3</cmath>
+
<cmath>\left(\frac{10a^2-5a+1}{b^2-5b+10}\right)\left(\frac{10b^2-5b+1}{c^2-5c+10}\right)\left(\frac{10c^2-5c+1}{a^2-5a+10}\right) \le(abc)^3</cmath>
 
Therefore,
 
Therefore,
−
<cmath> \min\left(\frac{10a^2-5a+1}{b^2-5b+10},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right )\leq abc. </cmath>
+
<cmath> \min\left(\frac{10a^2-5a+1}{b^2-5b+10},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right )\le abc. </cmath>

Latest revision as of 20:23, 15 April 2018

Problem

Let $a$, $b$, $c$ be real numbers greater than or equal to $1$. Prove that \[\min{\left (\frac{10a^2-5a+1}{b^2-5b+10},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right )}\leq abc\]

Solution

Since $(a-1)^5\ge 0$, \[a^5-5a^4+10a^3-10a^2+5a-1\ge 0\] or \[10a^2-5a+1\le a^3(a^2-5a+10)\] Since $a^2-5a+10=\left( a-\dfrac{5}{2}\right)^2 +\dfrac{15}{4}>0$, \[\frac{10a^2-5a+1}{a^2-5a+10}\le a^3\] Also note that $10a^2-5a+1=10\left( a-\dfrac{1}{4}\right)^2+\dfrac{3}{8}> 0$, We conclude \[0\le \frac{10a^2-5a+1}{a^2-5a+10}\le a^3\] Similarly, \[0\le \frac{10b^2-5b+1}{b^2-5b+10}\le b^3\] \[0\le \frac{10c^2-5c+1}{c^2-5c+10}\le c^3\] So \[\left(\frac{10a^2-5a+1}{a^2-5a+10}\right)\left(\frac{10b^2-5b+1}{b^2-5b+10}\right)\left(\frac{10c^2-5c+1}{c^2-5c+10}\right)\le a^3b^3c^3\] or \[\left(\frac{10a^2-5a+1}{b^2-5b+10}\right)\left(\frac{10b^2-5b+1}{c^2-5c+10}\right)\left(\frac{10c^2-5c+1}{a^2-5a+10}\right) \le(abc)^3\] Therefore, \[\min\left(\frac{10a^2-5a+1}{b^2-5b+10},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right )\le abc.\]

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