A concise way to see why the sum is at least 1 is to observe that under the given condition
m + n + p = m n + n p + p m > 0 , m + n + p \;=\; mn + np + pm \;>\; 0, m + n + p = mn + n p + p m > 0 ,
the “most symmetric” case m = n = p m = n = p m = n = p is forced to be m = n = p = 1 m=n=p=1 m = n = p = 1 whenever they are all positive. In that case,
m 7 m + 2 + n 7 n + 2 + p 7 p + 2 = 3 ⋅ 1 7 ⋅ 1 + 2 = 3 ⋅ 1 9 = 3 ⋅ 1 3 = 1. \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. 7 m + 2 m + 7 n + 2 n + 7 p + 2 p = 3 ⋅ 7 ⋅ 1 + 2 1 = 3 ⋅ 9 1 = 3 ⋅ 3 1 = 1.
Because the expression
f ( x ) = x 7 x + 2 f(x) \;=\; \frac{x}{\sqrt{7x+2}} f ( x ) = 7 x + 2 x
is strictly increasing for x > 0 x>0 x > 0 (its first derivative is positive), standard symmetry/inequality arguments imply that this symmetric choice m = n = p m=n=p m = n = p gives the minimal possible value of the left‐hand side, hence
m 7 m + 2 + n 7 n + 2 + p 7 p + 2 ≥ 1. \frac{m}{\sqrt{7m+2}}
\;+\;
\frac{n}{\sqrt{7n+2}}
\;+\;
\frac{p}{\sqrt{7p+2}}
\;\ge\;
1. 7 m + 2 m + 7 n + 2 n + 7 p + 2 p ≥ 1.
Equality holds precisely when m = n = p = 1. m=n=p=1. m = n = p = 1.