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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?

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    $\begingroup$ Can you solve that by the given hint? $\endgroup$ – user Oct 20 '18 at 22:32
  • $\begingroup$ Yes. Already accepted your answer. :> $\endgroup$ – Page not found Oct 20 '18 at 22:35
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    $\begingroup$ Yes I see that but I was interested to know if you were completely fine with the solution! Well done, Bye. $\endgroup$ – user Oct 20 '18 at 22:36
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HINT

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

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

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  • $\begingroup$ I never heard of this but it looks amazing. I was able to solve this problem now (took approx. 2 min.) after spending several hours trying to apply Cauchy. -.- You are my hero. :=) $\endgroup$ – Page not found Oct 20 '18 at 22:33
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    $\begingroup$ @Pagenotfound It is a foundamental inequality, absolutely to know and use when possible! $\endgroup$ – user Oct 20 '18 at 22:38
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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} $

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    $\begingroup$ That's nice but Jensen seems more effective here :) $\endgroup$ – user Oct 20 '18 at 23:03
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    $\begingroup$ Yes, but I think this is more elementary. $\endgroup$ – marty cohen Oct 21 '18 at 0:02
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    $\begingroup$ Convexity is really a very elementary concept, I'm not sure what is the more elementary among them. $\endgroup$ – user Oct 21 '18 at 0:04
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    $\begingroup$ I absolutely love this! (because it involves the Cauchy-Schwarz inequality I desperately tried to apply ;-)) $\endgroup$ – Page not found Oct 21 '18 at 19:01

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