Since a, b, c form a triangle, the area of the triangle could be expressed in terms of three sides of the trianlge per Heron's formula.

A = \dfrac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}

= \dfrac{1}{4}\sqrt{-a^4-b^4-c^4+2a^2b^2+2b^2c^2+2a^2c^2}

Since the radicand of square root must be positive, a,b,c satisfy the following condition.

2a^2b^2+2b^2c^2+2a^2c^2 > a^4+b^4+c^4

In order for \sqrt{a},\sqrt{b}, \sqrt{c} to form a triangle. \sqrt{a},\sqrt{b}, \sqrt{c} must satisfy the same condition, that is

2ab+abc+2ac>a^2+b^2+c^2

which is the conclusion to prove for the problem.

According to the triangle inequality theorem, the sum of any two sides of a triangle is greater than the third side of the triangle, that is

a+b>c ⇒ ac+bc>c^2

b+c>a ⇒ab+ac>a^2

c+a>b ⇒ bc+ab>b^2

Addition of above three inequalities yields proof of the conclusion.

2ab+abc+2ac>a^2+b^2+c^2

which means \sqrt{a},\sqrt{b}, \sqrt{c} also form a triangle.