Difference between revisions of "2018 USAMO Problems/Problem 1"
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WLOG, we can scale down all variables such that the lowest one is <math>1</math>. WLOG, let this be <math>a=1</math>. | WLOG, we can scale down all variables such that the lowest one is <math>1</math>. WLOG, let this be <math>a=1</math>. | ||
− | We now have <math>1+b+c=4\sqrt[3]{bc}</math>, and we want to prove <math>2bc+2b+2c+4\ge 1+b^2+c^2.</math> Adding <math>2bc</math> to both sides and subtracting <math>2b+2c</math> gives us <math>4bc+4\ge 1+ (b+c)(b+c-2)</math>, or <math>4bc+3\ge (b+c)(b+c-2)</math>. Let <math>\sqrt[3]{bc}=x</math>. Now, we have <cmath>4x^3+3 \ge (4x-1)(4x-3)</cmath> <cmath>4x^3 - 16x^2 + 16x \ge 0</cmath> <cmath>4x^2 - 16 + 16 \ge 0</cmath> <cmath>4(x-2)^2 \ge 0</cmath> By the trivial inequality, this is always true, | + | We now have <math>1+b+c=4\sqrt[3]{bc}</math>, and we want to prove <math>2bc+2b+2c+4\ge 1+b^2+c^2.</math> Adding <math>2bc</math> to both sides and subtracting <math>2b+2c</math> gives us <math>4bc+4\ge 1+ (b+c)(b+c-2)</math>, or <math>4bc+3\ge (b+c)(b+c-2)</math>. Let <math>\sqrt[3]{bc}=x</math>. Now, we have <cmath>4x^3+3 \ge (4x-1)(4x-3)</cmath> <cmath>4x^3 - 16x^2 + 16x \ge 0</cmath> <cmath>4x^2 - 16 + 16 \ge 0</cmath> <cmath>4(x-2)^2 \ge 0</cmath> By the trivial inequality, this is always true. Since all these steps are reversible, the proof is complete. |
+ | ~SigmaPiE |
Latest revision as of 21:47, 6 July 2023
Contents
[hide]Problem 1
Let be positive real numbers such that . Prove that
Solution
WLOG let . Add to both sides of the inequality and factor to get:
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
Solution 3
Similarly to Solution 2, we will prove homogeneity but we will use that to solve the problem differently. Let . Note that , thus proving homogeneity.
WLOG, we can scale down all variables such that the lowest one is . WLOG, let this be . We now have , and we want to prove Adding to both sides and subtracting gives us , or . Let . Now, we have By the trivial inequality, this is always true. Since all these steps are reversible, the proof is complete. ~SigmaPiE