I came across a result in a control theory book (without proof), which states that:
Given two variables $x,z \in \mathbb{R}$ and four parameters $c_{1}, c_{2}, k_{1}, k_{2}$ with $c_{1}, c_{2} > 0$. A function $f(x,z)$ is defined as
\begin{align} f(x,z) &= -c_{1}x^{2} - x^{4} + z[x-k_{1}z - k_{2}x^{2}z + (c_{1}-\sin x)(-c_{1}x - x^{3} + z)] \end{align}
The book reads that if we choose $k_{1} > c_{2} + c_{1} + 1 + \dfrac{(c_{1}^{2}+c_{1}+1)^{2}}{2c_{1}}$ and $k_{2} \ge \dfrac{(c_{1}+1)^2}{4}$, then $f(x,z) \le -\dfrac{1}{2}c_{1}x^{2} - c_{2}z^{2}$.
I am trying to prove such result, but no luck. Now, I am trying to expand every term of $f(x,z)$ to apply the AM-GM inequality, but it does not go anywhere. Do you have any suggestion/hint on this problem? Thanks a lot!
I only came up with an answer which is close to the result given by the book. But I think it is fine because this is not a tight inequality. First, I will expand all terms in $f(x,z)$ and then apply the AM-GM inequality to them. After exploiting every term in $f(x,z)$, we have
\begin{align} f(x,z) = -c_{1}x^{2} - x^{4} - (k_{1} - c_{1} + \sin{x})z^{2} - k_{2}x^{2}z^{2} + xz - c_{1}^{2}xz - c_{1}x^{3}z + c_{1}xz\sin{x} + x^{3}z\sin{x}. \end{align}
Then applying the AM-GM inequality for two numbers $a, b$ in $\mathbb{R}$: $ab \le|ab| \le \dfrac{1}{2}(a^{2} + b^{2})$ on the sign undetermined terms in $f(x,z)$, we have
\begin{align} x^{3}z\sin{x} = (x^{2})(xz\sin{x}) \le \dfrac{x^{4}}{2} + \dfrac{x^{2}z^{2}\sin^2{x}}{2} \le \dfrac{x^{4}}{2} + \dfrac{x^{2}z^{2}}{2}. \end{align}
\begin{align} c_{1}xz\sin{x} = \dfrac{1}{2}c_{1}(2xz\sin{x}) \le \dfrac{1}{2}c_{1}(\dfrac{x^{2}}{2} + \dfrac{4z^{2}\sin^{2}{x}}{2}) \le \dfrac{1}{4}c_{1}x^{2} + c_{1}z^{2} \end{align}
\begin{align} xz \le \dfrac{1}{4}c_{1}x^{2} + \dfrac{z^{2}}{c_{1}} \end{align}
\begin{align} -c_{1}^{2}xz \le \dfrac{1}{4}c_{1}x^{2} + c_{1}^{3}z^{2} \end{align}
\begin{align} -c_{1}x^{3}z \le \dfrac{x^{4}}{2} + \dfrac{c_{1}^{2}x^{2}z^{2}}{2} \end{align}
Then plug to $f(x,z)$, we have
\begin{align} f(x,z) \le -\dfrac{1}{4}c_{1}x^{2} - (k_{1} - c_{1} + \sin{x})z^{2} - k_{2}x^{2}z^{2} + (c^{3} + c_{1} + \dfrac{1}{c_{1}})z^{2} + \dfrac{1}{2}(1 + c_{1}^{2})x^{2}z^{2} \quad (1) \end{align}
Applying the basic inequalities $a^2 + b^2 + c^2 \ge \dfrac{(a+b+c)^2}{3}$ and $a^2 + b^2 \ge \dfrac{(a+b)^2}{2}$ with $a,b,c > 0$, we have
\begin{align} c_{1}^{3} + c_{1} + \dfrac{1}{c_{1}} = \dfrac{c_{1}^{4} + c_{1}^{2} + 1}{c_{1}} \ge \dfrac{(c_{1}^{2} + c_{1} + 1)^{2}}{3c_{1}} \end{align}
\begin{align} 1 + c_{1}^{2} \ge \dfrac{(1+c_{1})^2}{2} \end{align}
Therefore, in order to fulfill $(1)$ it is sufficient to choose $k_{1}$ and $k_{2}$ satisfying
\begin{align} f(x,z) \le -\dfrac{1}{4}c_{1}x^{2} - (k_{1} - c_{1} - 1 - \dfrac{(c_{1}^{2} + c_{1} + 1)^{2}}{3c_{1}})z^{2} - (k_{2} - \dfrac{(1+c_{1})^{2}}{4})x^{2}z^{2} \end{align}
Hence, by choosing \begin{align} k_{1} > 1 + c_{1} + c_{2} + \dfrac{(c_{1}^{2} + c_{1} + 1)^{2}}{3c_{1}} \end{align}
\begin{align} k_{2} \ge \dfrac{(1+c_{1})^{2}}{4}, \end{align}
we have
\begin{align} f(x,z) \le -\dfrac{1}{4}c_{1}x^{2} - c_{2}z^{2}. \end{align}