Proof of a linear algebra lemma for Cohn-Vossen's theorem For the proof of Cohn-Vossen's rigidity theorem I need to prove the next lemma (can be found in Montiel-Ros's Curves and Surfaces page 218):

If $\Phi$ and $\Psi$ are two definite self-adjoint endomorphisms of a Euclidean vector
  plane and $det \Phi = det \Psi$. Then, $det (\Phi + \Psi) \leq 0$
  and that equality occurs if and only if $\Phi = - \Psi$.

The proof says the following:

If  $det(\Phi +\Psi) > 0$, the endomorphism $\Phi+\Psi$ would be definite, say, positive definite. Now, we take $\{ e_1, e_2 \}$ a basis of the plane diagonalizing $\Phi$, one has $$ \langle \Phi(e_i), e_i \rangle + \langle \Psi(e_i), e_i \rangle > 0, \;\;\;\; i = 1,2.$$
  Consequently, $$ det \Phi = \langle \Phi(e_1), e_1 \rangle \langle \Phi(e_2), e_2 \rangle \stackrel{(1)}{>} \langle \Psi(e_1), e_1 \rangle \langle \Psi(e_2), e_2 \rangle \geq \langle \Psi(e_1), e_1 \rangle \langle \Psi(e_2), e_2 \rangle - \langle \Psi(e_1), e_2 \rangle^2 \stackrel{(2)}{=} det \Psi $$ which gives a contradiction to our hypothesis. Therefore, $det( \Phi + \Psi ) \leq 0$.

Here, I don't understand inequality (1) and equality (2). 
Moreover, the lemma says the following:

In the above inequality, equality occurs if and only if $\Phi = -
 \Psi$.

The proof says the following:

Following the reasoning above, if equality holds, there wouldbe at least a non-null vector in the kernel of $\Phi + \Psi$. Let $\{u_1, u_2 \}$ be a basis diagonalizing $\Phi + \Psi$, that is, such that $$ \Phi(u_1) + \Psi(u_1) = 0 \;\;\;\; and \;\;\;\; \Phi(u_2) + \Psi(u_2) = \lambda u_2, \;\;\; \lambda \in \mathbb{R}. $$
  From the first equality we deduce that $$ \langle \Phi(u_1), u_1 \rangle = - \langle \Psi(u_1), u_1 \rangle \;\;\;\; and \;\;\;\; \langle \Phi(u_1), u_2 \rangle = - \langle \Psi(u_1), u_2 \rangle, $$
  which together with the facts that $det \Phi = det \Psi$ and that $\Phi$ and $\Psi$ are definite, gives the equality $$ \langle \Phi(u_2), u_2 \rangle \stackrel{(3)}{=} - \langle \Psi(u_2), u_2 \rangle, $$
  implying $\lambda = 0$. Thus, in this case, $\Phi = - \Psi$.

Of this latest part I don't understand the equality (3).
Could you show me why these equations hold?
 A: Once I fixed a crucial typo, some things became clearer. Since we're assuming $\Phi+\Psi$ is positive definite, we have $\langle (\Phi+\Psi)e_i,e_i\rangle > 0$ for $i=1,2$, so
$\langle\Phi(e_i),e_i\rangle > -\langle\Psi(e_i),e_i\rangle$ for $i=1,2$. "Therefore," so to speak,
$$\langle\Phi(e_1)e_1\rangle\langle\Phi(e_2),e_2\rangle>\langle\Psi(e_1),e_1\rangle\langle\Psi(e_2),e_2\rangle,$$
establishing inequality (1). This is the proof the authors intended, but of course it's wrong unless we are assuming $\Psi$ is negative definite here, so that both right-hand sides are positive and we can multiply the inequalities.
Equality (2) is just the computation of $\det\Psi$ using a matrix representation with respect to the basis $\{e_1,e_2\}$.
Now that we've narrowed things down to assuming that $\Phi$ is positive definite and $\Psi$ is negative definite, let's look at (3). This is also following from a determinant computation: 
\begin{align*}
\det\Phi&=\langle \Phi(u_1),u_1\rangle \langle \Phi(u_2),u_2\rangle - \langle \Phi(u_1),u_2\rangle^2\\
\det\Psi &= \langle \Psi(u_1),u_1\rangle\langle \Psi(u_2),u_2\rangle -\langle \Psi(u_1),u_2\rangle^2.
\end{align*}
Since $\det\Phi=\det\Psi$, substituting the first two equalities, we get
$$\langle \Phi(u_1),u_1\rangle \langle \Phi(u_2),u_2\rangle - \langle \Phi(u_1),u_2\rangle^2 = -\langle \Psi(u_1),u_1\rangle \langle \Phi(u_2),u_2\rangle - \langle \Psi(u_1),u_2\rangle^2 = \langle \Psi(u_1),u_1\rangle\langle \Psi(u_2),u_2\rangle -\langle \Psi(u_1),u_2\rangle^2,$$
and so $\langle\Psi(u_2),u_2\rangle = -\langle \Phi(u_2),u_2\rangle$, as they claimed.
Let me reiterate that the lemma is false as stated. If we assume both $\Phi$ and $\Psi$ are positive definite, then we in fact should conclude that $\det(\Phi-\Psi)\le 0$ with equality holding iff $\Phi=\Psi$.  If we assume (as apparently these authors meant to) that $\Phi$ is positive definite and $\Psi$ is negative definite, then we conclude that $\det(\Phi+\Psi)\le 0$ with equality holding iff $\Phi=-\Psi$.
A: I just had a quick chat with Prof. Montiel and it seems to be a typo in the text. Essentially, you should substitute $det (\Phi + \Psi)$ by $det (\Phi - \Psi)$. 
This was the change that one could spot in the proof by Chern (page 87) that I pointed out in the comments. The difference between the proofs amounts to the fact that Montiel-Ros was taken from an article he didn't remember (he mentionned two mathematicians: the Spanish Santaló and a German called something like "Fechlossen").
I think that reusing the valuable answer by @TedShifrin you can now reconstruct what the lemma should look like and more importantly, you will be able to use it in the proof of Cohn-Vossen's theorem.
Let me know if this works out for the proof. 
