Prove $f(-\frac12) \le \frac{3}{16}$ if all roots of $f(x) = x^4 - x^3 + a x + b$ are real

Question

Let $a, b$ be real numbers such that all roots of $f(x) := x^4 - x^3 + ax + b$ are real. Prove that $f(-1/2) \le 3/16$.

The question was posted recently which was closed, due to missing of contexts etc.

My attempt: $f(-1/2) \le 3/16$ is equivalent to $a \ge 2b$.

Let $x_1, x_2, x_3, x_4$ be the real roots of $f(x)$. By Vieta, we have \begin{align*} x_1 + x_2 + x_3 + x_4 &= 1, \\ x_1 x_2 + x_1x_3 + x_1x_4 + x_2 x_3 + x_2x_4 + x_3x_4 &= 0, \\ x_1x_2x_3 + x_1x_2x_4 + x_1 x_3 x_4 + x_2x_3 x_4 &= -a, \\ x_1x_2x_3x_4 &= b. \end{align*}

The problem becomes:

If $x_1, x_2, x_3, x_4$ are real numbers such that $x_1 + x_2 + x_3 + x_4 = 1$ and $x_1 x_2 + x_1x_3 + x_1x_4 + x_2 x_3 + x_2x_4 + x_3x_4 = 0$, prove that $$x_1x_2x_3 + x_1x_2x_4 + x_1 x_3 x_4 + x_2x_3 x_4 + 2x_1x_2x_3x_4 \le 0.$$

This is true (e.g. verified by Mathematica). Is there a nice proof for it (or the original problem)?

Answer 1

Suppose $b= a/2+c$ where $c>0$. Then we have $$\underbrace{x^4-x^3+ax+{a\over 2}}_{p(x)} =-c$$ This means that graph of $p$ and line $y=-c<0$ have at least two different common points (if it has only one then $p(x)=(x-d)^4-c$ for some real $d$, this case is not possible) so $p$ has exactly two minimums with negative value, say at $x_1$ and $x_3$ where $x_1<x_3$, so $p(x_1)<0$ and $p(x_3)<0$ i.e. $$x_i^4-x_i^3+ax_i+{a\over 2} <0\;\;\;(*)$$ for $i=1,3$. Since $x_1,x_3$ satisfies also $p'(x_i)=0$ we have $$a=-4x_i^3+3x_i^2\;\;\;(**)$$ so we get plugging $(**)$ in inequality $(*)$ for $i=1,3$: $$-6x_i^4+3x_i^2<0\implies x_i^2>{1\over 2}$$

If we subtract equations $(**)$, we get $$(x_1-x_3)\Big(4(x_1^2+x_1x_3+x_3^2)-3(x_1+x_3)\Big)=0$$ so $$4(x_1^2+x_1x_3+x_3^2)=3(x_1+x_3)$$ and from here we get $$2(x_1^2+x_3^2)+2(x_1+x_3)^2=3(x_1+x_3)$$ which means that $$ 2+2t^2<3t$$ where $t=x_1+x_3$. But this inequality does not have a solution so we have a contradiction.

Answer 2

A proof for the equivalent formulation with Vieta's equations can be given by using Newton's identities which "help in guessing" but are not mandatory, see below.

Let's first write the definitions, using OP's formulation:

If $x_1, x_2, x_3, x_4$ are real numbers such that $e_1 = p_1 = x_1 + x_2 + x_3 + x_4 = 1$ and $e_2 = x_1 x_2 + x_1x_3 + x_1x_4 + x_2 x_3 + x_2x_4 + x_3x_4 = 0$, prove that $$e_3 + 2e_4 = x_1x_2x_3 + x_1x_2x_4 + x_1 x_3 x_4 + x_2x_3 x_4 + 2x_1x_2x_3x_4 \le 0.$$

Here, as in the link, $e_k$ are the elementary symmetric polynomials and $p_k$ is the k-th power sum.

We have the following, using Newton's identities:

$$ 3 e_3 = e_2 p_1 - e_1 p_2 + p_3 \\ 4 e_4 = e_3 p_1 - e_2 p_2 + e_1 p_3 - p_4 $$

For illustration, writing this fully expanded, the first line reads: $$ 3 (x_1x_2x_3 + x_1x_2x_4 + x_1 x_3 x_4 + x_2x_3 x_4) = \\ (x_1 x_2 + x_1x_3 + x_1x_4 + x_2 x_3 + x_2x_4 + x_3x_4)(x_1 + x_2 + x_3 + x_4)\\ - (x_1 + x_2 + x_3 + x_4)(x_1^2 + x_2^2 + x_3^2 + x_4^2) \\ + (x_1^3 + x_2^3 + x_3^3 + x_4^3) $$ It is a tedious but elementary task to check that this equality holds.

Now the questions asks for the values of $e_3 + 2 e_4$. Using the two Newton's identities this can be written as:

$$ 2(e_3 + 2 e_4) = \frac{2 + p_1}{3} [e_2 p_1 - e_1 p_2 + p_3] - e_2 p_2 + e_1 p_3 - p_4 $$

Note that so far no use of conditions given in the question has been made. The last result, if written down fully expanded, gives an utterly lengthy identity decomposition of $e_3 + 2 e_4$, which however could again be verified by elementary calculation. If someone had "guessed" this, no need would have arised to use Newton's identities.

Using the given conditions $e_1 = p_1 = 1$ and $e_2 = 0$, the sought value becomes

$$ 2(e_3 + 2 e_4) = [- p_2 + p_3 ] + p_3 - p_4 = -p_2 + 2 p_3 - p_4 $$

Expanding this gives

$$ 2(e_3 + 2 e_4) = - (x_1^2 + x_2^2 + x_3^2 + x_4^2) + 2 (x_1^3 + x_2^3 + x_3^3 + x_4^3) - (x_1^4 + x_2^4 + x_3^4 + x_4^4) \\ = - \Big[ x_1^2 (1-x_1)^2 + x_2^2 (1-x_2)^2 + x_3^2 (1-x_3)^2 + x_4^2 (1-x_4)^2 ) \Big] \le 0 $$

where the last conclusion follows since all terms are quadratic.

Indeed, equality is reached under the task's conditions only for $(x_1,x_2,x_3,x_4) = (0,0,0,1)$ and permutations thereof. $\qquad \Box$

Answer 3

(Becuase there is a "pre-calculus" tag, I will use calculus as little as possible in this argument for the original problem, since a couple of points can only be sufficiently refined by applying the derivative function.)

We might start with the function $ \ h(x) \ = \ x^4 - x^3 \ \ , \ $ which has a triple zero at the origin and its fourth (real) zero at $ \ x \ = \ 1 \ \ ; $ because $ \ h(x) \ > \ 0 \ $ for $ \ x \ < \ 0 \ $ and $ \ x \ > \ 1 \ \ , \ $ we must have $ \ h(x) \ < \ 0 \ $ in the interval $ \ (0 \ , \ 1) \ \ . \ $ We find $ \ \mathbf{h \left( -\frac12 \right) \ = \ \frac{3}{16} } \ \ , \ \ h \left( -\frac14 \right) \ = \ \frac{5}{256} \ \ , \ \ h \left( \frac14 \right) \ = \ \frac{3}{256} \ \ , \ \ h \left( \frac12 \right) \ = \ -\frac{1}{16} \ \ , $ $ h \left( \frac34 \right) \ = \ -\frac{27}{256} \ \ . \ $ (This last is in fact the global minimum of the function.) So the "bottom" of the function curve is rather "flat" with a modest "dip" in the interval $ \ \left( \frac12 \ , \ 1 \right) \ \ . $

Modifying this to $ \ g(x) \ = \ x^4 - x^3 + ax \ \ , \ $ has the effect of "tilting the floor" of the function curve. We will want to lower the portion of the curve "to the left" of the $ \ y-$axis while raising the portion "to the right" by taking $ \ a \ > \ 0 \ : \ $ this will create a "turning point" to the left of the origin and also to its right, since, in the neighborhood of the origin, $ \ g(x) \ $ is very close to just $ \ ax \ \ . $ At the $ \ x-$coordinates listed above, the function values are now

$$ \ \mathbf{g \left( -\frac12 \right) \ = \ \frac{3}{16} - \frac{a}{2} } \ \ , \ \ g \left( -\frac14 \right) \ = \ \frac{5}{256} - \frac{a}{4} \ \ , \ \ g \left( \frac14 \right) \ = \ \frac{3}{256} + \frac{a}{4} \ \ , $$ $$ g \left( \frac12 \right) \ = \ -\frac{1}{16} + \frac{a}{2} \ \ , \ \ g \left( \frac34 \right) \ = \ -\frac{27}{256} + \frac{3a}{4} \ \ . \ $$

We still have a zero at $ \ x \ = \ 0 \ \ , \ $ but it is now possible to have zeroes in $ \ \left( -\frac12 \ , \ -\frac14 \right) \ $ and $ \ \left( \frac14 \ , \ \frac34 \right) \ \ . $ The zero at $ \ x \ = \ 1 \ $ of $ \ h(x) \ $ "moves" to $ \ x \ < \ 1 \ $ since $ \ g(1) \ = \ a \ \ . $

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EDIT (8/6) -- If we bring in differentiation to investigate the number of turning points of the function, we have $ \ g'(x) \ = \ 4x^3 - 3x^2 + a \ \ , \ $ which yields the "depressed" cubic polynomial $ \ \mathcal{G}(x) \ = \ t^3 \ - \ \frac{3}{16}·t \ + \ \frac14·\left(a - \frac18 \right) \ \ . $ In order for $ \ g(x) \ $ to have three turning points, the discriminant must be $$ \Delta \ \ = \ \ 4·\left( -\frac{3}{16} \right)^3 \ + \ \frac{27}{4^2}·\left(a - \frac18 \right)^2 \ \ = \ \ \frac{27}{64}·a·\left(a - \frac14 \right) \ \ < \ \ 0 \ \ , $$ which is true in the interval $ \ 0 \ < \ a \ < \ \frac14 \ \ $ (in agreement with comments to the original post).

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We see, however, that there is a further limit to the magnitude of $ \ a \ $ in order to avoid raising the turning point near $ \ x \ = \ \frac34 \ $ above the $ \ x-$axis. So we must have $ \ -\frac{27}{256} + \frac{3a}{4} \ \le \ 0 $ $ \Rightarrow \ 0 \ \le \ a \ \le \ \sim \frac{9}{64} \ \approx \ 0.141 \ . \ $ (A more precise value turns out to be $ \ 0.148 \ \ . \ ) $ This gives us $ \ \frac{3}{16} \ \ge \ g \left( -\frac12 \right) \ \ge \ \sim \frac{3}{16} - \frac{9/64}{2} \ \ . $

We can then make a "vertical shift" by amending the function expression to $ \ f(x) \ = \ x^4 - x^3 + ax + b \ \ . \ $ We note that for $ \ a \ > \ 0 \ \ , \ \ b \ < \ 0 \ \ $ that the Rule of Signs will admit the possibility of as many as three positive real zeroes and one negative one; with $ \ b \ > \ 0 \ \ , \ $ we may have two of each. (Having $ \ a \ < \ 0 \ $ only allows a total of two "sign changes" for $ \ f(x) \ $ and $ \ f(-x) \ \ , \ $ which is consistent with our result from examining the derivative function.)

However, the turning points are rather "shallow", so $ \ | \ b \ | \ $ must be fairly small; otherwise, two of the consecutive turning points will be on the same side of the $ \ x-$axis, which will leave us with only two intercepts. For the purpose of answering the original question, we do not need to consider the effect of lowering the curve $ \ ( \ b \ < \ 0 \ ) \ \ , \ $ since we already have $ \ g \left( -\frac12 \right) \ = \ \frac{3}{16} - \frac{a}{2} \ \le \ \frac{3}{16} \ \ . \ $ The remaining question is whether raising the curve $ \ ( \ b \ > \ 0 \ ) \ $ can overcome the amount by which $ \ h \left( -\frac12 \right) \ $ has been reduced.

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EDIT (8/6) [replacing the "pre-calculus" argument] -- The complication in finding the limit of the amount of "vertical shift" that will still permit $ \ f(x) \ $ to have four $ \ x-$intercepts (four real zeroes) is that the turning points are asymmetrical; we would really prefer not to have to locate the relative minima of $ \ f(x) \ $ and evaluate them to find the limit on $ \ b \ \ . $ Instead, we will construct a symmetrical function by using the value of $ \ a \ $ that is in the center of the permissible interval, $ \ \mathfrak{G}(x) \ = \ x^4 - x^3 + \frac18·x \ \ . \ $ It is symmetrical about $ \ x \ = \ \frac14 \ \ , \ $ producing $$ \left(u + \frac14 \right)^4 \ - \ \left(u + \frac14 \right)^3 \ + \ \frac18·\left(u + \frac14 \right) \ \ = \ \ u^4 \ - \ \frac38·u^2 \ + \ \frac{5}{256} \ \ , $$ which factors as $ \ \left(u^2 - \frac{1}{16} \right)·\left(u^2 - \frac{5}{16} \right) \ \ , $ so there are indeed four real zeroes. The extrema are found from $$ \ \frac{d}{du} \ \left[ \ \left(u^2 - \frac{1}{16} \right)·\left(u^2 - \frac{5}{16} \right) \ \right] \ \ = \ \ 4·u·\left(u^2 - \frac{3}{16} \right) \ \ = \ \ 0 \ \ . $$ Hence, there is a maximum at $ \ u \ = \ 0 \ $ and two minima at $ \ u \ = \ \pm \frac{\sqrt3}{4} \ \ $ at which the minimum value for $ \ \mathfrak{G}(x) \ $ is given as $ \ \left( \pm \frac{\sqrt3}{4} \right)^4 - \frac38·\left( \pm \frac{\sqrt3}{4} \right)^2 + \frac{5}{256} \ = \ -\frac{4}{256} \ = \ -\frac{1}{64} \ \ . $

So $ \ \mathfrak{G}(x) \ + \ \frac{1}{64} $ is raised just to the point where its minima are tangent to the $ \ x-$axis, that is, at which its four distinct real zeroes become two real double zeroes. (By a similar argument, we cannot "lower" $ \ \mathfrak{G}(x) \ $ by more than the "height" of its maximum at $ \ x \ = \ \frac14 \ \ , \ $ which is $ \ \frac{5}{256} \ \ . \ $ Thus, for all four zeroes of $ \ f(x) \ $ to be real, we have at most $ \ 0 \ < \ a \ < \ \approx 0.148 \ $ and $ \ -\frac{5}{256} \ \le \ b \ \le \ \frac{1}{64} \ \ . \ $ (As we've said, the range for $ \ b \ $ is often much narrower.)

The function value at $ \ x \ = \ -\frac12 \ $ is then $ \ \large{ \mathbf{f \left( -\frac12 \right) \ \le \ \frac{3}{16} - \frac{a}{2} + b } } \ \ , \ $ with (generally) $ \ | \ b \ | \ \le \ \frac{a}{8} \ \ . \ $

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[For illustrative purposes, I kept my "pre-calculus" argument here; it turns out that it is fairly accurate:

With $ \ 0 \ < \ a \ < \ \frac{9}{64} \ \ $ and $ \ b \ > \ 0 \ \ , \ $ we cannot then "raise" the curve much more than to place $ \ f \left( -\frac18 \right) \ = \ \frac{9}{4096} - \frac{a}{8} + b \ = \ 0 \ \ , \ $ which is roughly where a turning point is located; this gives us $ \ b \ \approx \ \frac{a}{8} \ \ , \ $ so $ \ 0 \ < \ b \ < \ \frac{9}{512} \ \ . \ $ The function value for $ \ f(x) \ $ at $ \ x \ = \ -\frac12 \ \ $ is then no larger than about $$ f \left( -\frac12 \right) \ \ = \ \ \frac{3}{16} \ - \ \frac{9/64}{2} \ + \ \frac{9}{512} \ < \ \frac{3}{16} \ \ . \ ] $$