Need a hint on how to solve this inequality I want to show that, whenever $\frac{u_1^2}{p^2}+\frac{u_2^2}{q^2} \leq 1$ and $\frac{v_1^2}{p^2}+\frac{v_2^2}{q^2} \leq 1$, then
$$
  \frac{(\lambda \, u_1 + (1 - \lambda) \, v_1)^2}{p^2} + \frac{(\lambda \, u_2 + (1 - \lambda) \, v_2)^2}{q^2} \leq 1
$$
for all $0 \leq \lambda \leq 1$.
My (failed) attempt: I tried to apply the Cauchy-Schwarz inequality and got
$$
\frac{(\lambda \, u_1 + (1 - \lambda) \, v_1)^2}{p^2} + \frac{(\lambda \, u_2 + (1 - \lambda) \, v_2)^2}{q^2} \leq
\left(\lambda^2 + (1-\lambda)^2\right)
\left(
  \frac{u_1^2}{p^2} + \frac{u_2^2}{q^2} + \frac{v_1^2}{p^2} + \frac{v_2^2}{q^2}
\right) \leq 2
$$
What is the correct way to approach this?

Using simpler variable names,
we are given
$\frac{a^2}{p^2}+\frac{c^2}{q^2}
\le 1$
and
$\frac{b^2}{p^2}+\frac{d^2}{q^2}
\le 1$.
$\begin{array}\\
\frac{(ra + (1 - r) b)^2}{p^2} + \frac{(rc + (1 - r)d)^2}{q^2} 
&=\frac{r^2a^2+2r(1-r)ab+(1-r)^2b^2}{p^2} + \frac{r^2c^2+2r(1-r)cd+(1-r)^2d^2}{q^2}\\ 
&=r^2(\frac{a^2}{p^2}+\frac{c^2}{q^2})+2r(1-r)(\frac{ab}{p^2}+\frac{cd}{q^2})+(1-r)^2(\frac{b^2}{p^2}+\frac{d^2}{q^2})\\
&\le r^2+2r(1-r)(\frac{ab}{p^2}+\frac{cd}{q^2})+(1-r)^2\\
\end{array}
$
If we can show that
$\frac{ab}{p^2}+\frac{cd}{q^2}
\le 1$,
the upper bound is
$r^2+2r(1-r)+(1-r)^2
=(r+(1-r))^2 
=1
$
and we are done.
But
$\begin{array}\\
(\frac{ab}{p^2}+\frac{cd}{q^2})^2
&=(\frac{a}{p}\frac{b}{p}+\frac{c}{q}\frac{d}{q})^2\\
&\le(\frac{a^2}{p^2}+\frac{c^2}{q^2})(\frac{b^2}{p^2}+\frac{d^2}{q^2})
\quad\text{by Cauchy-Schwarz}\\
&\le 1\\
\end{array}
$
