Difference between revisions of "2018 USAMO Problems/Problem 1"
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| + | 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> | ||
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| + | The last inequality is true by AM-GM. Since all these steps are reversible, the proof is complete. | ||
Revision as of 14:42, 30 December 2018
Problem 1
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.
