Archit's Challenge 7

Algebra Level 5

For real numbers $a,b,c,d,p,q,r,s$ with $b-d\geq 5$ , let the roots of the quartic polynomial $f(x) = x^4 + ax^3 + bx^2 + cx + d$ be $p,q,r$ and $s$ . Find the minimum value of

$(1+p^2)(1+q^2)(1+r^2)(1+s^2).$

The answer is 16.

2 solutions

Aryan Goyat
Nov 28, 2015

multiply f(i) with f(-i ) we get (1+d-b)(1+d-b)+(c-a)(c-a) if c=a, d-b=-5 we get minimum

Did the same way

Righved K - 5 years, 6 months ago

you must show c=a yeilds real roots. i will show it for you: let c=a,d=1,b=6.then: $x^4+ax^3+6x^2+ax+1=0$ divede by x squared: $(x+\dfrac{1}{x})^2+a(x+\dfrac{1}{x})+4=0$ $x+\dfrac{1}{x}=\dfrac{-a\pm\sqrt{a^2-16}}{2}$ $x=\dfrac{\dfrac{-a\pm\sqrt{a^2-16}}{2}\pm\sqrt{(\dfrac{-a\pm\sqrt{a^2-16}}{2})^2-4}}{2}$ there clearly exists a real a for which all 4 are real, we can take a=4 for real. so, include this in your solution as due to silly mistake i got answer 36..... i cant write a solution. @aryan goyat

Aareyan Manzoor - 5 years, 6 months ago

Could you elaborate upon the need to show c=a produces real roots?

Aditya Dhawan - 4 years, 7 months ago

Harsh Poonia
Jul 18, 2019

AM-GM Inequality $b=(p+q)(r+s)+pq+rs$ $(p+q)(r+s) \geq 2 \sqrt{pq} \cdot 2 \sqrt{rs} \text{ (Equality holds when p=q, r=s.)}$ $pq+rs \geq 2 \sqrt{pqrs} \text{ (Equality when pq=rs.)}$ $b \geq 6 \sqrt{pqrs}$ Now the minimum value of $b-d$ should be greater than or equal to $5$ , so $6 \sqrt{pqrs} - pqrs \geq 5 \implies 5 \geq pqrs \geq 1.$ $\displaystyle\prod_{\text{cyclic}} (1+p^2) \geq 16 pqrs \geq 16. \text{ Equality at p=q=r=s=1}$

Archit's Challenge 7

The answer is 16.

2 solutions

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