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Difference between revisions of "2018 USAJMO Problems/Problem 2"

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== Problem ==
 
== 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.</cmath>
 
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>
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== Video Solution ==
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https://www.youtube.com/watch?v=eKIiaYNWUzM&t=5s
  
 
==Solution 1==
 
==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>
 
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>
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By substituting <math>a+b+c=\sqrt[3]{abc}</math>, we get:
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By substituting <math>a+b+c=4\sqrt[3]{abc}</math>, we get:
 
<cmath>\frac{4a\sqrt[3]{abc}+bc}{2} \geq 2\sqrt[3]{a^2b^2c^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.

Latest revision as of 16:38, 5 August 2024

Problem

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

Video Solution

https://www.youtube.com/watch?v=eKIiaYNWUzM&t=5s

Solution 1

WLOG let $a \leq b \leq c$. Add $2(ab+bc+ca)$ to both sides of the inequality and factor to get: \[4(a(a+b+c)+bc) \geq (a+b+c)^2\] By substituting $a+b+c=4\sqrt[3]{abc}$, we get: \[\frac{4a\sqrt[3]{abc}+bc}{2} \geq 2\sqrt[3]{a^2b^2c^2}\] The last inequality is true by AM-GM. Since all these steps are reversible, the proof is complete.

Solution 2

WLOG let $a \geq b \geq c$. Note that the equations are homogeneous, so WLOG let $c=1$. Thus, the inequality now becomes $2ab + 2a + 2b + 4 \geq a^2 + b^2 + 1$, which simplifies to $2(a+b) + 3 \geq (a-b)^2$.

Now we will use the condition. Letting $x=a+b$ and $y=a-b$, we have $x+1=\sqrt[3]{16(x^2-y^2)} \implies 16y^2=-x^3+13x^2-3x-1$.

Plugging this into the inequality, we have $2x+3 \geq \frac{1}{16}(-x^3+13x^2-3x-1) \implies x^3-13x^2+35x+49 = (x-7)^2(x+1) \geq 0$, which is true since $x \geq 0$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Solution 3

https://wiki-images.artofproblemsolving.com//6/69/IMG_8946.jpg 

-srisainandan6

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
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