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| − | == Problem ==
| + | #redirect[[2018 USAMO Problems/Problem 1]] |
| − | 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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| − | ==Solution 1==
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| − | 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>
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| − | The last inequality is true by AM-GM. Since all these steps are reversible, the proof is complete.
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| − | ==Solution 2==
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| − | WLOG let <math>a \leq b \leq c</math>. Note that the equations are homogeneous, so WLOG let <math>c=1</math>.
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| − | Thus, the inequality now becomes <math>2ab + 2a + 2b + 4 \geq a^2 + b^2 + 1</math>, which simplifies to <math>4ab + 3 \geq (a-b)^2</math>.
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| − | Now we will use the condition. Letting <math>x=a+b</math> and <math>y=a-b</math>, we have
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| − | <math>x+1=\sqrt[3]{16(x^2-y^2)} \implies 16y^2=-x^3+13x^2-3x-1</math>.
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| − | 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>.
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| − | {{MAA Notice}}
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| − | ==See also==
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| − | {{USAJMO newbox|year=2018|num-b=1|num-a=3}}
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