inequality of arithmetic and geometric means

The Problem:
	Find all pairs c and d of real numbers such that all roots of the polynomials 6x2−24x−4c and x3+cx2+dx−8 are nonnegative real numbers. Answer. There exists a single pair of values for which all roots are nonnegative and real, namely c=−6,d=12 .

The solution:

When c=−6 and d=12 the polynomials become 6(x−2)2 and (x−2)3 , whence all five roots coincide at x=2 and are thereby nonnegative and real. We must now show that there are no further satisfactory pairs. Everybody's first step was to consider the quadratic f(x):=6x2−24x−4c and show that for its roots to be real we must have c≥−6 : either note that its discriminant is 242−4⋅6⋅(−4c)=96(6+c) , which is nonnegative exactly when c≥−6 ; or write f(x)=6(x−2)2−4(c+6) and note that its minimum is −4(c+6) at x=2 , which is below the x -axis when c≥−6 . (The requirement that the roots of f(x) be nonnegative is not required; this extra condition forces c≤0 , but we shall soon see that the given cubic is more restrictive.)

To analyze the cubic we have a choice of using calculus or using the inequality of arithmetic and geometric means.

The AM-GM inequality. If r,s , and t are the roots of the cubic we have

g(x):=x3+cx2+dx−8=(x−r)(x−s)(x−t).

Comparing coefficients yields r+s+t=−c and rst=8 . To apply the AM-GM inequality we must make use of the requirement that the roots be nonnegative; in fact, they are positive (because their product is 8). The inequality tells us that any three positive numbers r,s,t satisfy

r+s+t≥3rst−−−√3,

whence −c≥6 ; that is, c≤−6 . But we earlier determined that c≥−6 , from which we deduce that c=−6 . It quickly follows that r+s+t=6=3rst−−−√3 ; therefore r=s=t=2 and d=12 . Consequently, c=−6,d=12 is the only pair satisfying the given conditions.



A calculus argument. The submissions that avoided the AM-GM inequality all used calculus to some extent. Here is how Harrison argued. Because the second derivative is g′′(x)=6x+2c , the cubic curve y=g(x)=x3+cx2+dx−8 has a point of inflection at (−13c,227c3−13cd−8) (where the second derivative is zero). Because g′(−13c)=d−13c2 , the line tangent to the curve at its inflection point has the equation

y−(227c3−13cd−8)=(d−13c2)(x+13c), or

y=(d−13c2)x−(127c3+8);

it therefore intersects the y -axis where y=−(127c3+8) . Because we know that c≥−6 , we deduce that the y -intercept is at most (127(−6)3+8)=0 . But we know that for our cubic g(x) to have distinct positive roots, it must attain its relative maximum for some value of x between x=0 and the inflection point, as in the accompanying diagram. Moreover, this maximum must lie below the line that is tangent to the curve at its inflection point; but we know that the inflectional tangent must have a negative slope, and we have seen that it lies below the positive x -axis, thus preventing the the curve from meeting the positive x -axis to the left of the inflection point. We are therefore led to conclude that the inflectional tangent must coincide with the x -axis. This forces 127c3+8=0 and, therefore, c=−6 . Finally, the slope of the tangent, namely d−13c2 , must be 0, so that d=12 , as claimed.

The graph of a cubic curve that has distinct positive roots,
shown with its inflectional tangent.

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