3
$\begingroup$

I have to prove that for all $x,y,z>0$,

$$\left(\frac{x+y}{x+y+z}\right)^{0.5} + \left(\frac{x+z}{x+y+z}\right)^{0.5} + \left(\frac{z+y}{x+y+z}\right)^{0.5} \leq 6^{0.5}$$

using Cauchy-Schwarz inequality? How do I do that ?

I have to define an inner product but I do not what, and what are the vectors?

$\endgroup$
7
$\begingroup$

Something to notice is that $\left(\frac{x+y}{x+y+z}\right)+ \left(\frac{x+z}{x+y+z}\right) + \left(\frac{z+y}{x+y+z}\right)= 2 $. This suggests writing $\left(\frac{x+y}{x+y+z}\right)^{0.5} + \left(\frac{x+z}{x+y+z}\right)^{0.5} + \left(\frac{z+y}{x+y+z}\right)^{0.5} $ as the dot product of $\left(\left(\frac{x+y}{x+y+z}\right)^{0.5},\left(\frac{x+z}{x+y+z}\right)^{0.5},\left(\frac{z+y}{x+y+z}\right)^{0.5}\right)$ with $(1,1,1)$, since the sum of the squares of the three terms will be on the right-hand side. And sure enough, using the Cauchy-Schwarz inequality will immediately give the bound you seek.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.