Difference between revisions of "2018 USAJMO Problems/Problem 2"
Nukelauncher (talk | contribs) (Created page with "== Problem == Let <math>a,b,c</math> be positive real numbers such that <math>a+b+c=4\sqrt[3]{abc}</math>. Prove that <cmath>2(ab+bc+ca)+4\min(a^2,b^2,c^2)\ge a^2+b^2+c^2.</cm...") |
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==Solution 1== | ==Solution 1== | ||
− | WLOG let <math>a \leq b \leq c | + | WLOG let <math>a \leq b \leq c</math>. Add <math>2(ab+bc+ca)</math> to both sides of the inequality and factor to get: <cmath>4(a(a+b+c)+bc) \geq (a+b+c)^2</cmath> <cmath>\frac{4a\sqrt[3]{abc}+bc}{2} \geq 2\sqrt[3]{a^2b^2c^2}</cmath> |
− | Add <math>2(ab+bc+ca)</math> to both sides of the inequality and factor to get: <cmath>4(a(a+b+c)+bc) \geq (a+b+c)^2</cmath> <cmath>\frac{4a\sqrt[3]{abc}+bc}{2} \geq 2\sqrt[3]{a^2b^2c^2}</cmath> | ||
The last inequality is true by AM-GM. Since all these steps are reversible, the proof is complete. | The last inequality is true by AM-GM. Since all these steps are reversible, the proof is complete. | ||
+ | |||
+ | ==Solution 2== | ||
+ | WLOG let <math>a \leq b \leq c</math>. Note that the equations are homogeneous, so WLOG let <math>c=1</math>. | ||
+ | Thus, the inequality now becomes <math>2ab + 2a + 2b + 4 \geq a^2 + b^2 + 1</math>, which simplifies to <math>4ab + 3 \geq (a-b)^2</math>. | ||
+ | |||
+ | Now we will use the condition. Letting <math>x=a+b</math> and <math>y=a-b</math>, we have | ||
+ | <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> since <math>x \geq 2</math>. | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 11:50, 26 April 2018
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 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 |