# For some $w$,$x$,$y$,$z$ it holds that $w+x>y+z$. Does that mean that $\sqrt w+\sqrt x> \sqrt y+\sqrt z$ and the other way around?

For some $w$,$x$,$y$,$z$ it holds that $$w+x>y+z. \quad \quad (1)$$Does that mean that $$\sqrt w+\sqrt x> \sqrt y+\sqrt z \quad \quad (2)$$ and the other way around, so if (2) holds then (1) too?

Thanks!

• $$\sqrt{25} + \sqrt{25} > \sqrt{49} + \sqrt{4}$$ – MathematicsStudent1122 Jun 14 '16 at 17:07

No. Counter-examples found by computer search:

$$1 \implies 2 \text{ counter-example } (w, x, y, z) = (1, 6, 3, 3)$$ $$2 \implies 1 \text{ counter-example } (w, x, y, z) = (2, 2, 1, 3)$$

(1) does not imply (2). For example $1 + 9 > 6 + 3$ but $\sqrt{6} + \sqrt{3} > \sqrt{1} + \sqrt{9}$. To find this counterexample I generated 4-tuples of random single-digit integers $(w, x, y, z)$ between 1 and 9 until one was a counterexample.

Given this, there's no need to find a separate counterexample for (2) implies (1). If

$$w + x > y + z \textrm{ and } \sqrt{w} + \sqrt{x} < \sqrt{y} + \sqrt{z}$$,

then, by reversing the inequalities,

$$\sqrt{y} + \sqrt{z} > \sqrt{w} + \sqrt{x} \textrm{ and } y + z < w + x.$$