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Let us consider the quartic polynomial in $x$ \begin{equation} F(x) = (2 a p +2)x^4+ (6a(1-a)p^2+(6-12a)p-6)x^3 + p(2(a-2)(a-1)a p^2 + 3(5a^2-9a+2)p +12a-18)x^2 - p^2 ((a-2)(4a^2 …

In order to prove inequality (5) it is sufficient to prove two inequalities $$V = 6(1-a)ap^2+(6-12a)p-6 > 0, \qquad (6V)$$ and $$Z = 2(a-2)(a-1)ap^2+3(5a^2-9a+2)p+12a-18 > 0 \quad (6Z)$$ for all $x > … Due to Vieta's formulas and $ ap+1 >0$ we obtain the following inequalities: $$3a(1-a)p^2+(3-2a)p+1 < 0, \qquad (18B)$$ $$4(a - 1)(a - 2)ap^3+ 6(-a^2- 3a+2)p^2 -24ap-12 > 0. …

The existence of the root $x_{*} > 1$ was proved above. Now we prove the uniquenes of the root $x_{*} > 1$. Let us suppose that there exists another root $x_{2,*} > 1$. Without loss of generality …

5a^2-9a+2)p +12a-18)x^2 \nonumber \\ - p^2 ((a-2)(4a^2 - 7a +1)p +9a^2-25a+18)x \nonumber \\ + (a-2)(a-1)(2a-3)p^3 =0, \quad (1) \end{eqnarray} where real parameters $a$ and $p$ obey inequalities … This task appears in solving certain physical problem and here all inequalities have certain physical meanings. …

There is a problem (of physical origin) which needs an analytical solution or a hint. Let us consider the following real-valued function of two variables $y (t,a) = 4 \left(1 + \frac{t}{x(t,a)}\right …