maximum value of $a+b+c$ given $a^2+b^2+c^2=48$? How can i get maximum value of this  $a+b+c$ given $a^2+b^2+c^2=48$ by not  using AM,GM and lagrange multipliers .

The Cauchy-Schwarz-inequality yields 
$$|a + b + c|^2 \le (a^2 + b^2 + c^2)\cdot 3 = 144$$
and therefore $a + b + c \le 12$. Plugging in $a = b = c = 4$ shows that this value is actually the maximum.
Alternatively, you could use the convexity of the function $x \mapsto \sqrt{x}$. By Jensen's inequality we have
$$a + b + c \le |a| + |b| + |c| = 3\left(\frac{1}{3}\sqrt{a^2} + \frac{1}{3}\sqrt{b^2} + \frac{1}{3}\sqrt{c^2}\right) \le 3 \sqrt{\frac{1}{3}a^2 + \frac{1}{3}b^2 + \frac{1}{3}c^2} = 12$$
Again, plugging in $a = b = c = 4$ yields the result.
