Proving that the Gauss and mean curvature of a compact surface in a closed disk $D[a,r]$ satisfies $K \geq \frac{1}{r^2}$, $H \geq \frac{2}{r}$

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the hiphotesys of the exercise ask for at least one point with that propertie.

My solution:

Define $f:S\rightarrow \mathbb{R}$ by $f(p)=|p-a|^2$, since $p$ is continuous and $S$ compact i have a global maximum $p$ such that $df^{2}(p)\leq 0$ but since $p$ is a extremum for $f$ the normal of $f$ at $p$ pass throught $a$ and we can write $a=p+\lambda N(p)$, such that $|\lambda|\leq r$, so derivating two times $f$ using the maximum hiphothesys and the second fundamental formula of $S$, $<v,dN(p).v>$, we get

$|v|^2-\lambda<v,dN(p).v>\leq 0$ for all $v$ in the tangent plane of $S$ at $p$, $T_{p}S$

so, if we get to a orthonormal basis of $T_{p}S$ $\{e_{1},e_{2}\}$ by the orthogonaliation process, if needed, i can tell that

$|e_{i}|^2-\lambda<e_{i},dN(p).e_{i}>\leq 0$ and then using the principal curvatures $k_{i}$ we have

$1-\lambda<e_{i},k_{i}e_{i}> \leq 0$ for $i=1,2$

$\Rightarrow$ $\lambda k_{i} \geq 1 \Rightarrow k_{i} \geq \frac{1}{r}$

and so i could conclude the exercise.

i have some doubts about the signs involving in theses inequalities and that why i'm not sure about this demonstration, anyonce can see any gap in this?