Difference between revisions of "2018 USAJMO Problems/Problem 2"
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==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> | ||
− | By substituting <math>a+b+c=\sqrt[3]{abc}</math>, we get: | + | 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. |
Revision as of 14:11, 7 March 2024
Problem
Let be positive real numbers such that
. Prove that
Solution 1
WLOG let . Add
to both sides of the inequality and factor to get:
By substituting
, we get:
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
.
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 |