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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 $$f(\cos \alpha, \cos \beta, \cos \gamma) \ge 0,$$ where \begin{align*} f(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 $f(a, b, c) \ge 0$ 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.

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.

Some thoughts.

With the substitution $t = \frac15 + s$, the equation is written as $$20s^3 - 18s^2 + qs - r = 0, \tag{1}$$ 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*}

We need to prove that the cubic equation (1) has at least one real root $s \le 0$.

Currently, we can prove that (1) does not have three positive real roots (see below). (to be continued.)

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.

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 $$f(\cos \alpha, \cos \beta, \cos \gamma) \ge 0,$$ where \begin{align*} f(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 $f(a, b, c) \ge 0$ 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.

Solution.

With the substitution $t = \frac15 + s$, the equation is written as $$20s^3 - 18s^2 + qs - r = 0, \tag{1}$$ 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*}

We claim that the cubic equation (1) has at least one real root $s \le 0$. 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.

We are done.

Some thoughts.

With the substitution $t = \frac15 + s$, the equation is written as $$20s^3 - 18s^2 + qs - r = 0, \tag{1}$$ 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*}

We need to prove that the cubic equation (1) has at least one real root $s \le 0$.

Currently, we can prove that (1) does not have three positive real roots (see below). (to be continued.)

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.