I’ve recently been looking into applications of the Cauchy Schwarz inequality.
I’ve seen it stated in the form:
$(a_1^2 + a_2^2 + … + a_n^2)(b_1^2 + b_2^2 + … + b_n^2) \ge (a_1b_1 + a_2b_2 + … + a_nb_n)^2$
However I’m struggling to see how it could prove this cyclic sum inequality:
$$\sum_\text{cyc} \frac{x^2}{y + z} \ge \frac{(x+y+z)^2}{2(x+y+z)}$$
Where $xyz = 1$ and $x,y,z$ $\epsilon$ $\mathbb{R}$
Any help?
Best Answer
Assuming $x, y, z \geq 0$, the Cauchy-Schwarz inequality gives
$$ \begin{align*} &\left((y + z) + (z + x) + (x + y)\right)\left(\frac{x^2}{y + z} + \frac{y^2}{z + x} + \frac{z^2}{x + y}\right) \\ &\geq \left(\sqrt{y + z}\sqrt{\frac{x^2}{y + z}} + \sqrt{z + x}\sqrt{\frac{y^2}{z + x}} + \sqrt{x + y}\sqrt{\frac{z^2}{x + y}}\right)^2 \\ &= (x + y + z)^2 \end{align*} $$
and you can divide out $(y + z) + (z + x) + (x + y) = 2(x + y + z)$ on both sides to get your inequality.