Am I real?

Algebra Level 3

Let a , b , c ∈ R + a,b,c \in \mathbb R^+ be distinct numbers such that a + b + c = 1 a+b+c=1 .

Let α = min ( a 3 + a 2 b c , b 3 + b 2 c a , c 3 + c 2 a b ) \alpha = \text{min}(a^3 +a^2bc, b^3 + b^2ca, c^3 + c^2ab)

Then, what can you say about the roots of the equation

x 2 + x + 4 α = 0 x^2 + x + 4\alpha = 0

Non-real Real and equal Real and distinct Can't say

Raushan Sharma by Raushan Sharma , 20, India

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1 solution

Raushan Sharma
Apr 1, 2016

W . L . O . G W.L.O.G , we may assume a < b < c a<b<c . Then, α = a 3 + a 2 b c \alpha = a^3 + a^2bc

Now, discriminant of the given equation x 2 + x + 4 α = 0 x^2 + x + 4\alpha = 0 is 1 − 16 α 1-16\alpha

Now, α = a 3 + a 2 b c = a 2 ( a + b c ) < a 2 ( a + ( b + c ) 2 4 ) = a 2 ( a + ( 1 − a ) 2 4 ) = a 2 ( a + 1 ) 2 4 < 16 9 ⋅ 9 ⋅ 4 = 4 81 < 1 16 \alpha = a^3 + a^2bc = a^2(a + bc) < a^2(a + \frac{(b+c)^2}{4}) = a^2(a + \frac{(1-a)^2}{4}) = \frac{a^2(a+1)^2}{4} < \frac{16}{9 \cdot 9 \cdot 4} = \frac{4}{81} < \frac{1}{16}

Therefore, 16 α < 1 ⇒ 1 − 16 α > 0 16\alpha < 1 \Rightarrow 1 - 16\alpha > 0

Hence, the roots are real and distinct.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Why can't you have b = c b=c ?

James Wilson - 3 years, 4 months ago

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Nice solution, by the way. (+1)

James Wilson - 3 years, 4 months ago

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