Polynomial Inequality

Algebra Level 5

Let a a and b b be real numbers such that the equation x 3 − a x 2 + b x − a = 0 x^3-ax^2+bx-a=0 has three positive real roots. Find the minimum of 2 a 3 − 3 a b + 3 a b + 1 \dfrac{2a^3-3ab+3a}{b+1} to three decimal places


The answer is 15.588.

Department 8 by Department 8 , 27, Israel

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1 solution

Department 8
Feb 22, 2016

x 1 + x 2 + x 3 = a > 0 , x 2 x 3 + x 3 x 1 + x 1 x 2 = b > 0 , x 1 x 2 x 3 = a > 0 a 3 = x 1 + x 2 + x 3 3 ≥ x 1 x 2 x 3 3 = a 3 ⇒ a ≥ 3 3 , a 2 = ( x 1 + x 2 + x 3 ) 2 ≥ 3 ( x 2 x 3 + x 3 x 1 + x 1 x 2 ) = 3 b ⇒ a 2 + 3 b + 1 ≥ 3 , ⇒ 2 a 3 − 3 a b + 3 a b + 1 = 2 a ⋅ a 2 + 3 b + 1 − 3 a ≥ 3 a ≥ 9 3 , ( 2 a 3 − 3 a b + 3 a b + 1 ) m i n = 9 3 . \large{x_1+x_2+x_3=a>0,x_2x_3+x_3x_1+x_1x_2=b>0,x_1x_2x_3=a>0 \\ \frac{a}{3}=\frac{x_1+x_2+x_3}{3}\ge \sqrt[3]{x_1x_2x_3}=\sqrt[3]{a} \Rightarrow a\ge 3\sqrt{3},\\ a^2=(x_1+x_2+x_3)^2\ge 3(x_2x_3+x_3x_1+x_1x_2) =3b\Rightarrow \frac{a^2+3}{b+1}\ge 3,\\ \Rightarrow \frac{2a^3-3ab+3a}{b+1}=2a\cdot \frac{a^2+3}{b+1}-3a \ge 3a \ge 9\sqrt{3},\\ \left(\frac{2a^3-3ab+3a}{b+1}\right)_{min}=9\sqrt{3}.}

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Second line, where does 3 3 3 \sqrt{3} come from?

Peter Pan - 4 years, 8 months ago

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