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Difference between revisions of "2018 USAMO Problems/Problem 1"

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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.  
 
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
 
~SigmaPiE
 +
 +
{{USAMO newbox|year=2018|before=First Question|num-a=2}}

Latest revision as of 01:30, 3 December 2024

Problem 1

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


Solution

WLOG let $a \leq b \leq c$. Add $2(ab+bc+ca)$ to both sides of the inequality and factor to get: \[4(a(a+b+c)+bc) \geq (a+b+c)^2\] \[\frac{4a\sqrt[3]{abc}+bc}{2} \geq 2\sqrt[3]{a^2b^2c^2}\]

The last inequality is true by AM-GM. Since all these steps are reversible, the proof is complete.

Solution 2

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 $f(a,b,c)=a+b+c-4\sqrt[3]{abc}$. Note that $f(a,b,c)=f(ka,kb,kc)$, thus proving homogeneity.

WLOG, we can scale down all variables such that the lowest one is $1$. WLOG, let this be $a=1$. We now have $1+b+c=4\sqrt[3]{bc}$, and we want to prove $2bc+2b+2c+4\ge 1+b^2+c^2.$ Adding $2bc$ to both sides and subtracting $2b+2c$ gives us $4bc+4\ge 1+ (b+c)(b+c-2)$, or $4bc+3\ge (b+c)(b+c-2)$. Let $\sqrt[3]{bc}=x$. Now, we have \[4x^3+3 \ge (4x-1)(4x-3)\] \[4x^3 - 16x^2 + 16x \ge 0\] \[4x^2 - 16 + 16 \ge 0\] \[4(x-2)^2 \ge 0\] By the trivial inequality, this is always true. Since all these steps are reversible, the proof is complete. ~SigmaPiE

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