Difference between revisions of "2014 USAJMO Problems/Problem 1"
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==Problem== | ==Problem== | ||
| − | 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+ | + | 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\ge 0</math>, | |
| + | <cmath>a^5-5a^4+10a^3-10a^2+5a-1\ge 0</cmath> | ||
| + | or | ||
| + | <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}>0</math>, | ||
| + | <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}> 0</math>, | ||
| + | We conclude | ||
| + | <cmath>0\le \frac{10a^2-5a+1}{a^2-5a+10}\le a^3</cmath> | ||
| + | Similarly, | ||
| + | <cmath>0\le \frac{10b^2-5b+1}{b^2-5b+10}\le b^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)\le a^3b^3c^3</cmath> | ||
| + | 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) \le(abc)^3</cmath> | ||
| + | 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 )\le abc. </cmath> | ||
Latest revision as of 21:23, 15 April 2018
Problem
Let 
, 
, 
be real numbers greater than or equal to 
. Prove that ![\[\min{\left (\frac{10a^2-5a+1}{b^2-5b+10},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right )}\leq abc\]](http://latex.artofproblemsolving.com/d/8/3/d831c8a9952575c830570b51daa569147932a714.png)
Solution
Since 
,
![\[a^5-5a^4+10a^3-10a^2+5a-1\ge 0\]](http://latex.artofproblemsolving.com/8/d/1/8d16b37367862c61bdaa92ab8b358bd3ae792314.png)
or
![\[10a^2-5a+1\le a^3(a^2-5a+10)\]](http://latex.artofproblemsolving.com/9/6/2/9620c7a73a23a9359ec0b9331be565b3e5339d7e.png)
Since 
,
![\[\frac{10a^2-5a+1}{a^2-5a+10}\le a^3\]](http://latex.artofproblemsolving.com/c/b/b/cbb02345b94aaded316592aae2e44f6e31cdf4fd.png)
Also note that 
,
We conclude
![\[0\le \frac{10a^2-5a+1}{a^2-5a+10}\le a^3\]](http://latex.artofproblemsolving.com/6/2/a/62a52a8e8690b0377156cb77618450107ad01a99.png)
Similarly,
![\[0\le \frac{10b^2-5b+1}{b^2-5b+10}\le b^3\]](http://latex.artofproblemsolving.com/f/2/4/f245c4f919b41f87157256aa4d54bccefa303be0.png)
![\[0\le \frac{10c^2-5c+1}{c^2-5c+10}\le c^3\]](http://latex.artofproblemsolving.com/e/8/c/e8c96f7e12f5d488c184aef7a4330efd255f3956.png)
So ![\[\left(\frac{10a^2-5a+1}{a^2-5a+10}\right)\left(\frac{10b^2-5b+1}{b^2-5b+10}\right)\left(\frac{10c^2-5c+1}{c^2-5c+10}\right)\le a^3b^3c^3\]](http://latex.artofproblemsolving.com/d/5/5/d559194865912b70bce050be0f6cc5fc022ba8ba.png)
or
![\[\left(\frac{10a^2-5a+1}{b^2-5b+10}\right)\left(\frac{10b^2-5b+1}{c^2-5c+10}\right)\left(\frac{10c^2-5c+1}{a^2-5a+10}\right) \le(abc)^3\]](http://latex.artofproblemsolving.com/9/b/b/9bbf8de655fa815c036cf73a10471023371e84a9.png)
Therefore,
![\[\min\left(\frac{10a^2-5a+1}{b^2-5b+10},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right )\le abc.\]](http://latex.artofproblemsolving.com/6/0/9/609fc79a7a5ffbe220ef57d2c13ab8be0def34b2.png)
