11. (20 points) Given the quadratic functions
$$
y_{1}=a x^{2}+4 x+b \text { and } y_{2}=b x^{2}+4 x+a
$$

both have minimum values, denoted as $m$ and $n$ for $y_{1}$ and $y_{2}$, respectively.
(1) If $m+n=0$, prove that for any real number $x$, $y_{1}+y_{2} \geqslant 0$;
(2) If $m$ and $n$ are both greater than 0, and $m n=2$, let $M$ be the larger of $m$ and $n$. Find the minimum value of $M$.
Three, 11. (1) From the conditions, we have
$$
\begin{array}{l}
a>0, b>0, \\
m=\frac{4 a b-4^{2}}{4 a}=\frac{a b-4}{a}, n=\frac{a b-4}{b} .
\end{array}
$$

From \( m+n=0 \), we know
$$
\begin{array}{l}
\frac{a b-4}{a}+\frac{a b-4}{b}=(a b-4)\left(\frac{1}{a}+\frac{1}{b}\right)=0 \\
\Rightarrow a b=4 .
\end{array}
$$

From \( y_{1}+y_{2}=(a+b) x^{2}+8 x+(a+b) \), its discriminant is
$$
\Delta=16-4(a+b)^{2} \leqslant 64-4(2 \sqrt{a b})^{2}=0 \text {, }
$$

we know that for any real number \( x \), \( y_{1}+y_{2} \geqslant 0 \).
(2) From \( m n=2 \), we know
$$
\begin{array}{l}
\frac{(a b-4)^{2}}{a b}=2 \\
\Rightarrow(a b)^{2}-10 a b+16=0 \\
\Rightarrow a b=8 \text { or } 2 .
\end{array}
$$

Noting that \( m, n > 0 \).
Then \( a b=2 \) should be discarded.
Thus, \( a b=8 \). At this time, \( m=\frac{4}{a}, n=\frac{4}{b} \).
Therefore, \( M \geqslant \frac{4}{a}, M \geqslant \frac{4}{b} \)
$$
\Rightarrow M^{2} \geqslant \frac{4}{a} \times \frac{4}{b}=\frac{16}{a b}=2 \Rightarrow M \geqslant \sqrt{2} \text {. }
$$

When \( a=b=2 \sqrt{2} \), \( M=\sqrt{2} \).
Thus, the minimum value of \( M \) is \( \sqrt{2} \).