# 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?

• Can you solve that by the given hint? – user Oct 20 '18 at 22:32
• Yes I see that but I was interested to know if you were completely fine with the solution! Well done, Bye. – user Oct 20 '18 at 22:36

HINT

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

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

• 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. :=) – Page not found Oct 20 '18 at 22:33
• @Pagenotfound It is a foundamental inequality, absolutely to know and use when possible! – user Oct 20 '18 at 22:38

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

$$\begin{array}\\ (\frac{ab}{p^2}+\frac{cd}{q^2})^2 &=(\frac{a}{p}\frac{b}{p}+\frac{c}{q}\frac{d}{q})^2\\ &\le(\frac{a^2}{p^2}+\frac{c^2}{q^2})(\frac{b^2}{p^2}+\frac{d^2}{q^2}) \quad\text{by Cauchy-Schwarz}\\ &\le 1\\ \end{array}$$

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