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 21:45, 29 April 2014
Problem
Let ,
,
be real numbers greater than or equal to
. Prove that
Solution
Since ,
or
Since
,
Also note that
,
We conclude
Similarly,
So
or
Therefore,