Need a hint on how to solve this inequality

Question

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?

Answer 1

HINT

Let use convexity for $f(x)=x^2$, that is by Jensen's inequality

$$f(\lambda x+(1-\lambda)y) \le \lambda f(x)+(1-\lambda)f(y)$$

Answer 2

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} $