Difference between revisions of "2018 USAMO Problems/Problem 1"

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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.
  
==Soltion 2==
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==Solution 2==
 
https://wiki-images.artofproblemsolving.com//6/69/IMG_8946.jpg
 
https://wiki-images.artofproblemsolving.com//6/69/IMG_8946.jpg
 
-srisainandan6
 
-srisainandan6

Latest revision as of 15:55, 8 October 2020

Problem 1

Let $a,b,c$ be positive real numbers such that $a+b+c=4\sqrt[3]{abc}$. Prove that \[2(ab+bc+ca)+4\min(a^2,b^2,c^2)\ge a^2+b^2+c^2.\]


Solution

WLOG let $a \leq b \leq c$. Add $2(ab+bc+ca)$ to both sides of the inequality and factor to get: \[4(a(a+b+c)+bc) \geq (a+b+c)^2\] \[\frac{4a\sqrt[3]{abc}+bc}{2} \geq 2\sqrt[3]{a^2b^2c^2}\]

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

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