Again set $h=\sqrt G$ and fix $\theta$. For $0<\alpha<C^{-1/2}$ define
$$
f(r)=h(r,\theta)-\alpha\sin(\sqrt C\,r).
$$
Since $h''=-Kh$ and $K\leq C$,
$$
f''+Cf=(C-K)h\geq0,
\qquad f(0)=0,
\qquad f'(0)=1-\alpha\sqrt C>0.
$$
For $0<r<\pi/\sqrt C$, differentiation gives
$$
\frac d{dr}
\left(f'\sin(\sqrt C r)-\sqrt C f\cos(\sqrt C r)\right)
=(f''+Cf)\sin(\sqrt C r)\geq0.
$$
The expression in parentheses vanishes at zero, and hence
$$
\left(\frac f{\sin(\sqrt C r)}\right)'\geq0.
$$
Its limit at zero is $(1-\alpha\sqrt C)/\sqrt C>0$, so $f>0$. Letting $\alpha\uparrow C^{-1/2}$ yields the <one-dimensional Rauch comparison inequality>
$$
h(r,\theta)\geq\frac{\sin(\sqrt C r)}{\sqrt C}
\qquad(0\leq r<\pi/\sqrt C).
$$
It follows that
$$
\operatorname{Area}B(p,\varepsilon)
\geq\frac{2\pi}{C}\bigl(1-\cos(\sqrt C\,\varepsilon)\bigr).
$$
Now
$$
\frac{2(1-\cos x)}{x^2}\longrightarrow1
$$
as $x\to0$. Given $0<\gamma<1$, choose $x_0\in(0,\pi)$ such that $2(1-\cos x)/x^2\geq\gamma$ for $0<x\leq x_0$, and set $\varepsilon_0=x_0/\sqrt C$. Then every geodesic polar coordinate ball of radius $\varepsilon\leq\varepsilon_0$ satisfies
$$
\boxed{\operatorname{Area}B(p,\varepsilon)\geq\gamma\pi\varepsilon^2.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 25I / d / Solution

If $h''+Kh=0$, $h(0)=0$, $h'(0)=1$, $h>0$, and $K\leq C$ for $C>0$, then
$$
h(r)\geq\frac{\sin(\sqrt C r)}{\sqrt C}
\qquad(0\leq r<\pi/\sqrt C).
$$
Subtract a smaller multiple of the sine solution and use the monotonicity of its Wronskian with $h$.

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 Incoming links: One-dimensional Rauch comparison inequality