Ta có:

`P = (a + b + c)/(\sqrt{a(a + 3b)} + \sqrt{b(b + 3c)} + \sqrt{c(c + 3a)})`

`= [2(a + b + c)]/[2[(\sqrt{a(a + 3b)} + \sqrt{b(b + 3c)} + \sqrt{c(c + 3a)}]`

`= [2(a + b + c)]/(\sqrt{4a(a + 3b)} + \sqrt{4b(b + 3c)} + \sqrt{4c(c + 3a)})`

Áp dụng BĐT Cô - si, ta có:

`\sqrt{4a(a + 3b)} ≤ (4a + a + 3b)/2 = (5a + 3b)/2`

`\sqrt{4b(b + 3c)} ≤ (4b + b + 3c)/2 = (5b + 3c)/2`

`\sqrt{4c(c + 3a)} ≤ (4c + c + 3a)/2 = (5c + 3a)/2`

`⇔ \sqrt{4a(a + 3b)} + \sqrt{4b(b + 3c)} + \sqrt{4c(c + 3a)} ≤ (5a + 3b)/2 + (5b + 3c)/2 + (5c + 3a)/2 = (8a + 8b + 8c)/2 = 4(a + b + c)`

$\\$

`⇔ [2(a + b + c)]/(\sqrt{4a(a + 3b)} + \sqrt{4b(b + 3c)} + \sqrt{4c(c + 3a)}) ≥ [2(a + b + c)]/[4(a + b + c)] = 1/2`

Vậy GTNN của `P` là `1/2` khi: $\begin{cases} 4a = a + 3b\\4b = b + 3c\\4c = 3a + c \end{cases}$

             `⇔ a = b = c`