The constraint implies $x^2=z^2+1$, so $x\ne 0$ and its gradient $(2x,0,-2z)$ never vanishes on the feasible set. The Lagrange multiplier equations $\nabla f=\lambda \nabla g$ giveThus $\lambda =1$, $y=0$, $z=-1/2$, and feasibility gives $x=\pm \sqrt{5}/2$. The two stationary points and their values areThey are both global minima. Indeed, on the constraint, completing the square givesEquality holds only at the two displayed points. There are no other constrained stationary points, and the objective is unbounded above as $|y|\to \infty$.

The arithmetic-geometric mean inequality for the positive numbers $x$ and $a/x$ givesEquality holds precisely when $x=a/x$, hence at $x=\sqrt{a}$. Thus $\boxed{\underset{x>0}{\min }g(x)=\sqrt{a}}$, attained only at $\boxed{x=\sqrt{a}}$. Equivalently, the exact identity $g(x)-\sqrt{a}=(x-\sqrt{a}{)}^2/(2x)$ proves the same global bound.

Write $Y(t)=y(e^t)$, so $t=\log x$. The chain rule givesSubstitution into the Cauchy-Euler equation therefore gives $Y^{\prime \prime }+Y^{\prime }-2Y=0$. Its characteristic equation factors as $(r-1)(r+2)=0$, so $Y=C_1{e}^t+C_2{e}^{-2t}$. Returning to $x$ gives

The homogeneous characteristic equation is $r^2-4r+3=(r-1)(r-3)=0$, giving $e^x$ and $e^{3x}$. Since the forcing $e^{3x}$ already appears among these solutions, try a particular solution $y_p=Ax{e}^{3x}$. Differentiation and substitution giveso $A=1/2$. The general solution is

After division by $x>0$, the equation is linear with coefficient $-1/x$ on $y$. Its integrating factor is $1/x$, and the product rule givesUsing the antiderivative from the preceding part,The arbitrary constant multiplies the homogeneous solution $x$.

Let $I=\int e^x\sin xdx$. Two applications of integration by parts givewhere additive constants can be absorbed at the end. Solving for $I$ yieldsDifferentiating the answer gives $e^x\sin x$.

The number $K$ is fixed. By Big O notation, for each $i$ there are constants $C_i>0$ and $N_i$ such that $f_i(n)\le C_{i}g(n)$ when $n\ge N_i$. Take $N={\max }_i{N}_i$. For $n\ge N$, positivity givesThe constants in parentheses are independent of $n$, proving $\boxed{h=O(g),l=O(g^2)}$. If $K=1$, the second sum is empty and $l=0$, which satisfies the same upper bound.