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 .
 A: 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.
A: Observe that we can formulate this question in terms of linear algebra.  Note the condition that $a^2 + b^2 + c^2 = 48$ is the same as a vector with norm $\sqrt{48} = 4\sqrt{3}$.  The condition of maximizing $a+b+c$ is the same as maximizing the inner product of $\mathbb{1} = \left [ \begin{array}{ccc}
1&1&1\\
\end{array}\right ]$ with the vector $v =\left [ \begin{array}{ccc}
a&b&c\\
\end{array} \right ]$.  We know however by Cauchy-Schwarz that $\langle \mathbb{1}, v\rangle$ is maximized when $v$ is linearly dependent with $\mathbb{1}$. In other words we would like $v$ to be a multiple of $\mathbb{1}$ hence $a=b=c$, such that $3a^2 = 48$, or rather $a = 4$.
A: You can use the identity
$$(a+b+c)^2=3(a^2+b^2+c^2)-(a-b)^2-(b-c)^2-(c-a)^2$$
The right-hand side is clearly maximised when $a=b=c$ and then $a+b+c=12$
A: 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 $a=b=c=4$
A: By the arithmetic-quadratic mean
inequality,
$\frac{a+b+c}{3}
\le \sqrt{\frac{a^2+b^2+c^2}{3}}
= \sqrt{\frac{48}{3}}
=\sqrt{16}
=4
$
with equality iff
$a=b=c$.
Therefore
$a+b+c = 12$
and
$a=b=c = 4$.
