Prove Cauchy Schwarz, I have the two vectors (which i decided upon my understanding, feel free to correct me if I'm wrong)

Question

2/(x+y) + 2/(y+z) + 2/(z+x) >= 9/(x+y+z)

I am using the following vectors:

(x+y y+z z+x) and (1 1 1)

I obtain something resembling the aforementioned inequality, but maybe my choice of vectors is totally wrong. Let me know if I should post a picture of my work!

Thank you.

Answer 1

Hint: You already have a simple way, but if you must use Cauchy-Schwarz for vectors, consider for e.g. $A=[\frac1{\sqrt{x+y}}, \frac1{\sqrt{y+z}}, \frac1{\sqrt{z+x}}]$ and $B = [\sqrt{x+y}, \sqrt{y+z}, \sqrt{z+x}]$.

Now use CS: $(A\cdot B)^2 \leqslant ||A||^2||B||^2$

Answer 2

Let $\,s=x+y+z\,$ then by the AM-HM inequality for the positive $s-x, s-y, s-z \gt 0\,$:

$$ \frac{2(x+y+z)}{3} = \frac{(s-x)+(s-y)+(s-z)}{3} \ge \frac{3}{\dfrac{1}{s-x}+\dfrac{1}{s-y}+\dfrac{1}{s-z}} $$