Cfrgtkky

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}xzsin{x} + x^{3}zsin{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}zsin{x} = (x^{2})(xzsin{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}xzsin{x} = dfrac{1}{2}c_{1}(2xzsin{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}