Difference between revisions of "2018 USAJMO Problems/Problem 2"
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<math>x+1=\sqrt[3]{16(x^2-y^2)} \implies 16y^2=-x^3+13x^2-3x-1</math>. | <math>x+1=\sqrt[3]{16(x^2-y^2)} \implies 16y^2=-x^3+13x^2-3x-1</math>. | ||
− | Plugging this into the inequality, we have <math>2x+3 \geq \frac{1}{16}(-x^3+13x^2-3x-1) \implies x^3-13x^2+35x+49 = (x-7)^2(x+1) \geq 0</math>, which is true since <math>x \geq | + | Plugging this into the inequality, we have <math>2x+3 \geq \frac{1}{16}(-x^3+13x^2-3x-1) \implies x^3-13x^2+35x+49 = (x-7)^2(x+1) \geq 0</math>, which is true since <math>x \geq 0</math>. |
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 10:21, 17 April 2019
Contents
Problem
Let be positive real numbers such that . Prove that
Solution 1
WLOG let . Add to both sides of the inequality and factor to get:
The last inequality is true by AM-GM. Since all these steps are reversible, the proof is complete.
Solution 2
WLOG let . Note that the equations are homogeneous, so WLOG let . Thus, the inequality now becomes , which simplifies to .
Now we will use the condition. Letting and , we have .
Plugging this into the inequality, we have , which is true since .
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See also
2018 USAJMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |