Problem 33

Algebra Level 5

$\large \dfrac{5a^2-3ab+2}{a^2b-a^3}$

Let $a$ and $b$ be non-zero real numbers such that all the roots of the equation $ax^3-x^2+bx-1=0$ are positive reals.

Given that the minimum value of the expression above can be expressed as $H\sqrt{K}$ and the equality achieved when $a=\dfrac{A\sqrt{B}}{C}, \qquad b=\sqrt{S}$ where $H,K,A,B, C$ and $S$ are positive integers with $K,B,S$ being square-free numbers and $\gcd(A,C)=1$ .

Compute $H+K+A+B+C+S$ .

Vietnam TST.

This problem is in this Set .

The answer is 31.

1 solution

Kushal Dey
Jun 5, 2021

Let given expression be M. Let ax³-x²+bx-1=0=(x-p)(x-q)(x-r). Then, p+q+r=pqr=1/a and pq+qr+rp=b/a. 5a²-3ab+2=a²(5-3(Σpq)+2(Σp)²) =a³(5pqr-3(Σpq)(Σp)+2(Σp)³) =a³(2(Σp(p+q)(r+p))+(p+q)(q+r)(r+p)) a²b-a³=a³((Σpq)-pqr/(Σp))=a⁴(p+q)(q+r)(r+p). Thus, M= $\frac {1}{a}$ (1+2(Σ $\frac {p}{q+r}$ )). Notice $\frac {1}{a²}$ =(p+q+r)²= $\frac {(p+q+r)³}{pqr}$ . Now it can be easily shown by AM-GM-HM inequality that, (p+q+r)³>=27pqr and Σ $\frac {p}{q+r}$ >= $\frac 32$ . Thus M>=3 $\sqrt 3$ *4 =12 $\sqrt 3$ . Equality achieved when p=q=r. Thus p³=3p => p= $\sqrt 3$ => a= $\frac {\sqrt 3}{9}$ , b= $\sqrt 3$ .
Thus A=1,B=3,C=9,S=3,H=12,K=3. Sum=31.

Problem 33

Vietnam TST.

This problem is in this Set .

The answer is 31.

1 solution

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