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> | ||
| + | |||
| + | == Video Solution == | ||
| + | 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> <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> |
| − | + | 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> | ||
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== | ==Solution 2== | ||
| − | WLOG let <math>a \ | + | WLOG let <math>a \geq b \geq c</math>. Note that the equations are homogeneous, so WLOG let <math>c=1</math>. |
| − | Thus, the inequality now becomes <math>2ab + 2a + 2b + 4 \geq a^2 + b^2 + 1</math>, which simplifies to <math> | + | Thus, the inequality now becomes <math>2ab + 2a + 2b + 4 \geq a^2 + b^2 + 1</math>, which simplifies to <math>2(a+b) + 3 \geq (a-b)^2</math>. |
Now we will use the condition. Letting <math>x=a+b</math> and <math>y=a-b</math>, we have | Now we will use the condition. Letting <math>x=a+b</math> and <math>y=a-b</math>, we have | ||
<math>x+1=\sqrt[3]{16(x^2-y^2)} \implies 16y^2=-x^3+13x^2-3x-1</math>. | <math>x+1=\sqrt[3]{16(x^2-y^2)} \implies 16y^2=-x^3+13x^2-3x-1</math>. | ||
| − | Plugging this into the inequality, we have <math>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</math> since <math>x \geq | + | Plugging this into the inequality, we have <math>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</math>, which is true since <math>x \geq 0</math>. |
{{MAA Notice}} | {{MAA Notice}} | ||
| + | |||
| + | ==Solution 3== | ||
| + | https://wiki-images.artofproblemsolving.com//6/69/IMG_8946.jpg | ||
| + | -srisainandan6 | ||
==See also== | ==See also== | ||
{{USAJMO newbox|year=2018|num-b=1|num-a=3}} | {{USAJMO newbox|year=2018|num-b=1|num-a=3}} | ||
Latest revision as of 16:38, 5 August 2024
Problem
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)
Video Solution
https://www.youtube.com/watch?v=eKIiaYNWUzM&t=5s
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)
By substituting , we get:
![\[\frac{4a\sqrt[3]{abc}+bc}{2} \geq 2\sqrt[3]{a^2b^2c^2}\]](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
WLOG let 
. Note that the equations are homogeneous, so WLOG let 
.
Thus, the inequality now becomes 
, which simplifies to 
.
Now we will use the condition. Letting 
and 
, we have
![$x+1=\sqrt[3]{16(x^2-y^2)} \implies 16y^2=-x^3+13x^2-3x-1$](http://latex.artofproblemsolving.com/5/f/d/5fd2cb1f8d03f37bfb01a08a541767ed4ee0f6c0.png)
.
Plugging this into the inequality, we have 
, which is true since 
.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
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 | ||
