Difference between revisions of "2018 USAMO Problems/Problem 1"
(→Solution 1) |
(→Soltion 2) |
||
| (2 intermediate revisions by 2 users not shown) | |||
| Line 3: | Line 3: | ||
| − | ==Solution | + | == Solution == |
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> | 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> | ||
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== | ||
| + | https://wiki-images.artofproblemsolving.com//6/69/IMG_8946.jpg | ||
| + | -srisainandan6 | ||
Revision as of 15:55, 8 October 2020
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
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.
Solution 2
https://wiki-images.artofproblemsolving.com//6/69/IMG_8946.jpg -srisainandan6
