2018 USAJMO Problems/Problem 2
Revision as of 00:32, 20 April 2018 by 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...")
Problem
Let 
be positive real numbers such that ![$a+b+c=4\sqrt[3]{abc}$](http://latex.artofproblemsolving.com/b/6/9/b69ff720e6d0801983832367327a018e7afce9ca.png)
. Prove that ![\[2(ab+bc+ca)+4\min(a^2,b^2,c^2)\ge a^2+b^2+c^2.\]](http://latex.artofproblemsolving.com/7/0/4/7043a170ac5535465a5a1b4daccb1f1de3043280.png)
Solution 1
WLOG let 
Add 
to both sides of the inequality and factor to get: ![\[4(a(a+b+c)+bc) \geq (a+b+c)^2\]](http://latex.artofproblemsolving.com/1/9/c/19cf714c622f8daf8c0d9b14c6d11e7d67e976f0.png)
![\[\frac{4a\sqrt[3]{abc}+bc}{2} \geq 2\sqrt[3]{a^2b^2c^2}\]](http://latex.artofproblemsolving.com/9/7/6/97633532a495f02b4209186344c3ba7896ad3cc4.png)
The last inequality is true by AM-GM. Since all these steps are reversible, the proof is complete.
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 | ||
