Inspired by Deepanshu Gupta

Algebra Level 5

f ( x ) = x 4 + a x 3 + b x 2 + c x + 1 f(x)=x^4+ax^3+bx^2+cx+1

If f ( x ) f(x) has at least one real root, for real numbers a , b a,b and c c , find the minimal value of a 2 + b 2 + c 2 a^2+b^2+c^2 . Write your answer in the form p q \frac{p}{q} for co-prime positive integers p p and q q , and enter p + q p+q .

If you come to the conclusion that no minimum is attained, enter 666.

Inspiration .


The answer is 7.

Otto Bretscher by Otto Bretscher , 65, USA

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1 solution

Otto Bretscher
Nov 1, 2015

If x 4 + a x 3 + b x 2 + c x + 1 = 0 x^4+ax^3+bx^2+cx+1=0 for some x x , then ( x 4 + 1 ) 2 = ( a x 3 + b x 2 + c x ) 2 (x^4+1)^2=(ax^3+bx^2+cx)^2 ≤ ( a 2 + b 2 + c 2 ) ( x 6 + x 4 + x 2 ) \leq (a^2+b^2+c^2)(x^6+x^4+x^2) by Cauchy-Schwarz, so that a 2 + b 2 + c 2 ≥ ( x 4 + 1 ) 2 x 6 + x 4 + x 2 = ( x 2 + 1 / x 2 ) 2 x 2 + 1 + 1 / x 2 ≥ 4 3 a^2+b^2+c^2\geq \frac{(x^4+1)^2}{x^6+x^4+x^2}=\frac{(x^2+1/x^2)^2}{x^2+1+1/x^2}\geq \boxed{\frac{4}{3}} . The last step is an easy exercise in calculus or inequalities (make x 2 + 1 + 1 / x 2 = t ≥ 3 ) x^2+1+1/x^2=t\geq {3}) .

The minimum is attained when x = 1 x=1 and a = b = c = − 2 / 3 a=b=c=-2/3 , for example.

19 Helpful 1 Interesting 2 Brilliant 0 Confused

Moderator note:

Oh wow, that's an interesting way to pull out the term a 2 + b 2 + c 2 a^ 2 + b^2 + c ^2 . And we can also generalize it to the inspiration problem.

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