We want to prove: [MATH]\sqrt{a}+\sqrt{b}+\sqrt{c}≥\sqrt{a+b+c} \hspace2ex[/MATH] (1)
Note [MATH]a, b, c ≥0[/MATH], so they can be written as [MATH]a=x^2, b=y^2, c=z^2[/MATH]
Also, we can choose [MATH]x, y, z[/MATH] to be: [MATH]x≥0, y≥0, z≥0[/MATH]Now, e.g. [MATH]\sqrt{a}=\sqrt{x^2}=x \hspace1ex[/MATH] since [MATH]x≥0[/MATH]
So expression (1) becomes: [MATH]x+y+z≥\sqrt{x^2+y^2+z^2}[/MATH]Since all terms are [MATH]≥0[/MATH], this inequality holds:
[MATH]\Leftrightarrow (x+y+z)^2≥x^2+y^2+z^2[/MATH][MATH]\Leftrightarrow (x+y+z)^2-(x^2+y^2+z^2)≥0 \hspace2ex[/MATH] (2)
When you expand and simplify this expression, you should be able to use the fact that [MATH]x≥0, y≥0, z≥0[/MATH] to show that it is [MATH]≥0[/MATH].