2018 USAJMO Problems/Problem 2: Difference between revisions

Latest revision as of 20:35, 2 December 2024

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-== Problem ==
+#redirect[[2018 USAMO Problems/Problem 1]]
-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.</cmath>
-==Solution 1==
-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.
-{{MAA Notice}}
-==See also==
-{{USAJMO newbox|year=2018|num-b=1|num-a=3}}