If α, β, γ are the positive real roots of the equation x3-6 x2+a x+b=0 then the minimum value of (3/α+1)+(3/β+1)+(3/γ+1) is

Question

If α, β, &gamma; are the positive real roots of the equation x3-6 x2+a x+b=0 then the minimum value of (3/α+1)+(3/β+1)+(3/&gamma;+1) is (A) 1 (B) 3 (C) 9 (D) 27

Tardigrade Admin · Accepted Answer

A.M. ≥ H.M. ⇒ ((3/α+1)+(3/β+1)+(3/&gamma;+1)/3) ≥ (3/(α+1)3+(β+1)3+(&gamma;+1/3) (3/α+1)+(3/β+1)+(3/&gamma;+1) ≥ (3 ⋅ 3 ⋅ 3/(α+β+&gamma;)+3) ≥ (27/9) (α+β+&gamma;=6) ⇒ (3/α+1)+(3/β+1)+(3/&gamma;+1) ≥ 3

Q. If α,β,γ are the positive real roots of the equation x3−6x2+ax+b=0 then the minimum value of α+13​+β+13​+γ+13​ is

1

3

9

27

A.M. ≥ H.M. ⇒3α+13​+β+13​+γ+13​​≥3α+1​+3β+1​+3γ+1​3​ α+13​+β+13​+γ+13​≥(α+β+γ)+33⋅3⋅3​≥927​(α+β+γ=6) ⇒α+13​+β+13​+γ+13​≥3

Q. If $α, β, γ$ are the positive real roots of the equation $x^{3} - 6 x^{2} + a x + b = 0$ then the minimum value of $\frac{3}{α + 1} + \frac{3}{β + 1} + \frac{3}{γ + 1}$ is