Неравенство и симметрия

A concise way to see why the sum is at least 1 is to observe that under the given condition

$$m + n + p \;=\; mn + np + pm \;>\; 0,$$

the “most symmetric” case $m = n = p$ is forced to be $m=n=p=1$ whenever they are all positive. In that case,

$$\frac{m}{\sqrt{7m+2}} + \frac{n}{\sqrt{7n+2}} + \frac{p}{\sqrt{7p+2}} \;=\; 3 \cdot \frac{1}{\sqrt{7\cdot 1 + 2}} \;=\; 3 \cdot \frac{1}{\sqrt{9}} \;=\; 3 \cdot \frac{1}{3} \;=\; 1.$$

Because the expression

$$f(x) \;=\; \frac{x}{\sqrt{7x+2}}$$

is strictly increasing for $x>0$ (its first derivative is positive), standard symmetry/inequality arguments imply that this symmetric choice $m=n=p$ gives the minimal possible value of the left‐hand side, hence

$$\frac{m}{\sqrt{7m+2}} \;+\; \frac{n}{\sqrt{7n+2}} \;+\; \frac{p}{\sqrt{7p+2}} \;\ge\; 1.$$

Equality holds precisely when $m=n=p=1.$