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 .

We prove if $a+b+c=r$ then $a^2+b^2+c^2>\frac{r^2}{3}$ and equality is reached only when $a=b=c$.
Suppose the minimum is achieved with values $a,b,c$ and we do not have $a=b=c$. then without loss of generality we have $a\neq b$ let $a+b=m$ and write $a$ as $\frac{m+n}{2}$ and $b$ as $\frac{m-n}{2}$. Then the sum of the squares is $\frac{m^2+n^2}{2}+c^2$ which is greater than what we would get with $\frac{m}{2},\frac{m}{2},c$. Hence the minimum is reached when $a=b=c$.
When $r=12$ we get the minimum value of $a^2+b^2+c^2$ is $\frac{144}{3}=48$. When $r$ is larger $\frac{r^2}{3}$ is larger than $48$. Hence the maximum value of $a+b+c$ is $12$ and is reached only when $a=b=c=4$
