Difference between revisions of "2014 USAJMO Problems/Problem 1"
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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+1},\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>, | |
| − | + | <cmath>a^5-5a^4+10a^3-10a^2+5a-1\geqslant 0</cmath> | |
| + | or | ||
| + | <cmath>10a^2-5a+1\leqslant a^3(a^2-5a+10)</cmath> | ||
| + | Since <math>a^2-5a+10=\left( a-\dfrac{5}{2}\right) +\dfrac{15}{4}\geqslant 0</math>, | ||
| + | <cmath> \frac{10a^2-5a+1}{a^2-5a+10}\leqslant a^3 </cmath> | ||
| + | Also note that <math>10a^2-5a+1=10\left( a-\dfrac{1}{4}\right)+\dfrac{3}{8}\geqslant 0</math>, | ||
| + | We conclude | ||
| + | <cmath>0\leqslant\frac{10a^2-5a+1}{a^2-5a+10}\leqslant a^3</cmath> | ||
| + | Similarly, | ||
| + | <cmath>0\leqslant\frac{10b^2-5b+1}{b^2-5b+10}\leqslant b^3</cmath> | ||
| + | <cmath>0\leqslant\frac{10c^2-5c+1}{c^2-5c+10}\leqslant 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> | ||
| + | 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> | ||
| + | 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> | ||
Revision as of 20:45, 29 April 2014
Problem
Let 
, 
, 
be real numbers greater than or equal to 
. Prove that ![\[\min{\left (\frac{10a^2-5a+1}{b^2-5b+1},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right )}\leq abc\]](http://latex.artofproblemsolving.com/3/a/d/3ad6f9ab816469889e5b8452f203c0e9b01647f4.png)
Solution
Since 
,
![\[a^5-5a^4+10a^3-10a^2+5a-1\geqslant 0\]](http://latex.artofproblemsolving.com/9/e/5/9e5d6fae3dfc6dfb543a0b37b5dec0b12068a14d.png)
or
![\[10a^2-5a+1\leqslant a^3(a^2-5a+10)\]](http://latex.artofproblemsolving.com/b/d/d/bddb7b52ee2ad720467d4b48c75fb50e32bfe26f.png)
Since 
,
![\[\frac{10a^2-5a+1}{a^2-5a+10}\leqslant a^3\]](http://latex.artofproblemsolving.com/f/0/6/f06c19fdf574548beb22059aeae7c1cd8dc2590c.png)
Also note that 
,
We conclude
![\[0\leqslant\frac{10a^2-5a+1}{a^2-5a+10}\leqslant a^3\]](http://latex.artofproblemsolving.com/3/0/0/30070d5deefd86d2b66d8a99e7e3396d4753eef4.png)
Similarly,
![\[0\leqslant\frac{10b^2-5b+1}{b^2-5b+10}\leqslant b^3\]](http://latex.artofproblemsolving.com/c/e/4/ce49e33c8bc0a8fa71cf1ef8d878d7def4dceb63.png)
![\[0\leqslant\frac{10c^2-5c+1}{c^2-5c+10}\leqslant c^3\]](http://latex.artofproblemsolving.com/c/5/5/c5570e69180e66136ba7286050089206fd62a2a1.png)
So ![\[\left(\frac{10a^2-5a+1}{a^2-5a+10}\right)\left(\frac{10b^2-5b+1}{b^2-5b+10}\right)\left(\frac{10c^2-5c+1}{c^2-5c+10}\right)\leqslant a^3b^3c^3\]](http://latex.artofproblemsolving.com/a/6/e/a6e524b9387b1a96f2b280aa5f14470778ce7b5c.png)
or
![\[\left(\frac{10a^2-5a+1}{b^2-5b+10}\right)\left(\frac{10b^2-5b+1}{c^2-5c+10}\right)\left(\frac{10c^2-5c+1}{a^2-5a+10}\right) \leqslant(abc)^3\]](http://latex.artofproblemsolving.com/0/c/5/0c5e7426c448f4f4d3f33f6399bcfb14a2f49209.png)
Therefore,
![\[\min\left(\frac{10a^2-5a+1}{b^2-5b+10},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right )\leq abc.\]](http://latex.artofproblemsolving.com/5/8/6/586aa2ebfc9ff2507752ab79760d05d894abe799.png)
