Smallest root of a degree 3 polynomial

Question

Is it true that the smallest root $t$ of the polynomial

$$ 20 t^3 - 30 t^2 + (12 - 4 \cos^2 \alpha - 4 \cos^2 \beta - 4 \cos^2 \gamma) t + \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma - 2 \cos \alpha \cos \beta \cos \gamma - 1 $$

is always no greater than $\frac{1}{5}$ for all $\alpha, \beta, \gamma \in [0,\pi]$ with $\alpha + \beta \geq \gamma$, $\alpha + \gamma \geq \beta$, $\beta + \gamma \geq \alpha$, and $\alpha + \beta + \gamma \leq 2\pi$?

Thanks a lot for any helpful answer.

Answer 1

This proof is similar to River Li's, but it requires no machine calculation. Also, we will show more, namely that the original polynomial has three real roots, the smallest of which is at most $$u:=\frac{5-\sqrt{15}}{10}=0.11270166\dotsc$$ This estimate is sharp, because in case of $\alpha=\beta=\gamma=\pi/2$, the polynomial is $$20t^3-30t^2+12t-1=20\left(t-\frac{1}{2}\right)\left(t-\frac{5-\sqrt{15}}{10}\right)\left(t-\frac{5+\sqrt{15}}{10}\right).$$ We turn to the proof. With the notation $$Q:=\frac{\cos^2\alpha+\cos^2\beta+\cos^2\gamma}{3} \qquad\text{and}\qquad R:=\cos\alpha\cos\beta\cos\gamma,$$ the polynomial in the original post equals $$p(t):=20t^3-30t^2+12(1-Q)t+(3Q-2R-1).$$ The coefficients of this cubic polynomial are $$a:=20,\qquad b:=-30,\qquad c:=12(1-Q),\qquad d:=3Q-2R-1,$$ hence its discriminant equals \begin{align*} \Delta\ := \ & 18 a b c d - 4 b^3 d + b^2 c^2 - 4 a c^3 - 27 a^2 d^2 \\ \ = \ & 2160 (1 + 12 Q + 3 Q^2 + 64 Q^3 - 60 Q R - 20 R^2). \end{align*} Here $2 QR\leq Q^2+R^2$, and also $R^2\leq Q^3$ by the AM-GM inequality, whence \begin{align*} \Delta/2160 \ & \geq \ 1 + 12 Q + 3 Q^2 + 64 Q^3 - 30 Q^2 - 50 R^2 \\ & \geq \ 1 + 12 Q - 27 Q^2 + 14 Q^3 \\ & = \ (Q-1)^2(1+14Q) \\ & \geq \ 0. \end{align*} So $p(t)$ has three real roots (counted with multiplicity), and the same is true of the shifted polynomial $$p(u+t)=20t^3-6\sqrt{15}t^2+6(1-2Q)t+\left(6\sqrt{\frac{3}{5}}Q-3Q-2R\right).$$ It remains to show that this new polynomial $p(u+t)$ has a nonpositive root $t$. If this is not the case, then $p(u+t)$ has three positive roots (counted with multiplicity), hence in particular $$1-2Q>0>6\sqrt{\frac{3}{5}}Q-3Q-2R.$$ However, this is a contradiction, because it would yield that $$Q^3\geq R^2>\left(3\sqrt{\frac{3}{5}}-\frac{3}{2}\right)^2 Q^2\geq\frac{1}{2}Q^2\geq Q^3.$$

Answer 2

Solution.

With the substitution $t = \frac15 + s$, the equation is written as $$f(s) := 20s^3 - 18s^2 + qs - r = 0, \tag{1}$$ where \begin{align*} q &:= \frac{12}{5} - 4 \cos^2 \alpha - 4 \cos^2 \beta - 4 \cos^2 \gamma, \\ r &:= - \frac{9}{25} - \frac15\cos^2 \alpha - \frac15 \cos^2 \beta - \frac15 \cos^2 \gamma + 2 \cos \alpha \cos \beta \cos \gamma. \end{align*}

First, let us prove that (1) has three real roots. It suffices to prove that discriminant of $f$ is non-negative, or equivalently $$g(\cos \alpha, \cos \beta, \cos \gamma) \ge 0,$$ where \begin{align*} g(a, b, c) &:= 64\,{a}^{6}+192\,{a}^{4}{b}^{2}+192\,{a}^{4}{c}^{2}+192\,{a}^{2}{b}^{4 }\\ &\qquad -156\,{a}^{2}{b}^{2}{c}^{2}+192\,{a}^{2}{c}^{4} +64\,{b}^{6}+192\,{b}^ {4}{c}^{2}\\ &\qquad +192\,{b}^{2}{c}^{4} +64\,{c}^{6}-540\,{a}^{3}bc -540\,a{b}^{3 }c\\ &\qquad -540\,ab{c}^{3}+9\,{a}^{4} +18\,{a}^{2}{b}^{2}+18\,{a}^{2}{c}^{2}+9\, {b}^{4}\\ &\qquad +18\,{b}^{2}{c}^{2}+9\,{c}^{4}+108\,{a}^{2}+108\,{b}^{2}+108\,{ c}^{2}+27. \end{align*} We can prove that $g(a, b, c) \ge 0$ for all real numbers $a, b, c$ (this is verified by Mathematica). The desired result follows.

Second, we prove that (1) does not have three positive real roots. Assume, for the sake of contradiction, that (1) has three positive real roots. Then we have $q > 0, r > 0$. However, we can prove that $\frac54 r + \frac{3}{16}q \le 0$ that is $$- \cos^2 \alpha - \cos^2 \beta - \cos^2 \gamma + \frac52 \cos \alpha \cos \beta \cos \gamma \le 0. \tag{2}$$ (2) is true since $$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma \ge 3\sqrt[3]{\cos^2 \alpha \cos^2 \beta \cos^2 \gamma}$$ $$\ge 3|\cos \alpha \cos \beta \cos \gamma| \ge \frac52 \cos \alpha \cos \beta \cos \gamma.$$ Contradiction. Thus, the claim is proved.

Thus, (1) has at least one real root $s \le 0$.

We are done.

Answer 3

Write $a,b,c$ instead of $\alpha,\beta,\gamma$. Let $p(t)$ be the polynomial in question. Let $$u:=\cos^2a+\cos^2b+\cos^2c,\quad v:=\cos a\,\cos b\,\cos c,$$ $$t_*:=\frac{1}{30} \left(15-\sqrt{15} \sqrt{4 u+3}\right).$$ Then $$0\le u\le3,\quad v\le(u/3)^{3/2},$$ $$\begin{aligned}p(t_*)&=\frac{(4 u+3)^{3/2}}{3 \sqrt{15}}-u-2 v \\ &\ge\frac{(4 u+3)^{3/2}}{3 \sqrt{15}}-u-2 (u/3)^{3/2} \\ &=:g(u)\ge g(3)=0 \end{aligned}$$ (the latter inequality holds because $u=3$ is the only critical point of $g$ and $g(0)>0$),
$$t_*>1/5\implies u<3/5,$$ $$\begin{aligned}p(1/5)&=\frac{1}{25} (5 u-50 v+9) \\ &\ge\frac{1}{25} (5 u-50(u/3)^{3/2}+9) \\ &=:h(u)\ge\min(h(0),h(3/5)) \\ &>0 \text{ if }0\le u<3/5 \end{aligned}$$ (the 2nd inequality in the latter display holds because $h$ is concave). So, $p(\min(t_*,1/5))\ge0$.

Also, $p(-\infty+)=-\infty<0$.

Thus, $p(t)=0$ for some real $t\le1/5$. $\quad\Box$

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Note that no conditions on $a,b,c$ (that is, on $\alpha,\beta,\gamma$) were needed or used here.