Cyclic Inequality!

We claim that $A = 2$ .

Let $f(x,y,z) = \dfrac{x}{\sqrt{y^2 + z^2}} + \dfrac{y}{\sqrt{z^2 + x^2}} + \dfrac{z}{\sqrt{x^2 + y^2}}$ .

We'll show that $f(x, y, z) > 2$ . Since $f(x, y, z) \to 2 \text{ as } x \to y \text{ and } z \to 0$ , the lower bound 2 is sharp. Without loss of generality, assume that $x \geq y \geq z$ . Since by the AM-GM Inequality, we have

$\dfrac{\sqrt{z^2 + x^2}}{\sqrt{y^2 + z^2}} + \dfrac{\sqrt{y^2 + z^2}}{\sqrt{z^2 + x^2}} \geq 2$

It suffices to show that

$f(x,y,z) > \dfrac{\sqrt{z^2 + x^2}}{\sqrt{y^2 + z^2}} + \dfrac{\sqrt{y^2 + z^2}}{\sqrt{z^2 + x^2}}$

or equivalently,

$\dfrac{z}{\sqrt{x^2 + y^2}} > \dfrac{\sqrt{z^2 + x^2} - x}{\sqrt{y^2 + z^2}} + \dfrac{\sqrt{y^2 + z^2} - y}{\sqrt{z^2 + x^2}}$

By simple algebra, this is easily seen to be equivalent to:

$\dfrac{z}{\sqrt{y^2 + z^2}(\sqrt{z^2 + x^2} + x)} + \dfrac{z}{\sqrt{z^2 + x^2}(\sqrt{y^2 + z^2} + y)} < \dfrac{1}{\sqrt{x^2 + y^2}} \text{ .... (1) }$

Since $\sqrt{y^2 + z^2} \geq \sqrt{2z^2} = \sqrt{2}z, \sqrt{z^2 + x^2} > x \text{ and } \sqrt{2}x \geq \sqrt{x^2 + y^2}$ , we have:

$\dfrac{z}{\sqrt{y^2 + z^2}(\sqrt{z^2 + x^2} + x)} < \dfrac{z}{\sqrt{2}z(x+x)} = \dfrac{1}{2\sqrt{2}x} \leq \dfrac{1}{2\sqrt{x^2 + y^2}}$

Thus to establish (1), it remains to show that:

$\dfrac{z}{\sqrt{z^2 + x^2}(\sqrt{y^2 + z^2} + y)} < \dfrac{1}{2\sqrt{x^2 + y^2}}$

or equivalently:

$\dfrac{2z}{\sqrt{y^2 + z^2} + y} < \sqrt{\dfrac{z^2 + x^2}{x^2 + y^2}}$

Since

$\dfrac{z^2 + x^2}{x^2 + y^2} = 1 - \dfrac{y^2 - z^2}{x^2 + y^2}$

which is a non-decreasing function of $x$ , we have:

$\dfrac{z^2 + x^2}{x^2 + y^2} \geq \dfrac{z^2 + y^2}{2y^2}$

and it suffices to show that

$\dfrac{\sqrt{z^2 + y^2}}{\sqrt{2}y} > \dfrac{2x}{\sqrt{z^2 + y^2} + y}$

or equivalently:

$z^2 + y^2 + y\sqrt{z^2 + y^2} \geq 2yz \text{ .... (2)}$

Since $y^2 + z^2 \geq 2yz$ , we have

$z^2 + y^2 + y\sqrt{z^2 + y^2} \geq 2yz + y\sqrt{2z^2} = (2 + \sqrt{2})yz > 2\sqrt{2}yz$

and thus (2) holds. This completes the proof.