Prove that $ \sum_1^3 x_i^4+\sum_1^3 x_i^6\le?$ when $\sum_1^3 x_i^2=3$ Prove that $\displaystyle \sum_1^3 x_i^4+\sum_1^3 x_i^6\le k$ when $\displaystyle\sum_1^3 x_i^2=3$.Find k.

$$\begin{align}
\sum_i^3x_i^4
&=\left(\sum_i^3x_i^2\right)^2-2\sum_{i,j\ne i}^3x_i^2xj^2\\
&=9-2\sum_{i,j\ne i}^3x_ixj\\
\sum_i^3x_i^6
&=\left(\sum_i^3x_i^2\right)\left(\sum_i^3x_i^4-\sum_{i,j\ne i}^3x_i^2x_j^2\right)+3\prod_i^3x_i^2\\
&=3\left(9-3\sum_{i,j\ne i}^3x_ixj\right)+3\prod_i^3x_i^2\end{align}$$
I can't reduce $\displaystyle \sum x_i^2x_j^2$ and $\displaystyle \prod x_i^2$
And tried using AM-GM which doesn;t give any upper bound instead the lower one.

NOTE: Use only AM-GM/Rearraangement/Chebyshev/Holder/Cauchy/Schur
 A: Note that
$$
\sum_{i=1}^nx_i^{2\lambda}\le\left(\sum_{i=1}^nx_i^2\right)^\lambda\tag{1}
$$
since expanding the right side (using the multinomial theorem) gives the left side plus other non-negative terms.
Furthermore, suppose that $x_1=\sqrt{n}$ and $x_i=0$ for $i\ne1$. Then we have equality in $(1)$. Thus, $(1)$ is sharp. That is,
$$
\sum_{i=1}^nx_i^2=n\implies\sum_{i=1}^nx_i^{2\lambda}\le n^\lambda\tag{2}
$$
is sharp.

Second Version of the Question
Using $(2)$, we have that
$$
\begin{align}
\sum_{i=1}^3x_i^4+\sum_{i=1}^3x_i^6
&\le3^2+3^3\\
&=36\tag{3}
\end{align}
$$
Note that equality in $(3)$ is attained if we let $x_1=\sqrt3$ and $x_2=x_3=0$. Thus, $(3)$ is as good as one can get.

Lower Bound
The upper bound is gotten by looking at the boundary cases, while the lower bound is gotten by using Hölder's Inequality.
Applying Hölder, we get
$$
\begin{align}
\sum_{i=1}^nx_i^21^{1-1/\lambda}
&\le\left(\sum_{i=1}^nx_i^{2\lambda}\right)^{1/\lambda}\left(\sum_{i=1}^n1\right)^{1-1/\lambda}\\
&=\left(\sum_{i=1}^nx_i^{2\lambda}\right)^{1/\lambda}\quad n^{1-1/\lambda}\tag{4}
\end{align}
$$
which implies
$$
n^{1-\lambda}\left(\sum_{i=1}^nx_i^2\right)^\lambda\le\sum_{i=1}^nx_i^{2\lambda}\tag{5}
$$
That is,
$$
\sum_{i=1}^nx_i^2=n\implies n\le\sum_{i=1}^nx_i^{2\lambda}\tag{6}
$$
for which equality holds if all $x_i=1$.

Therefore,
$$
\sum_{i=1}^nx_i^2=n\implies n\le\sum_{i=1}^nx_i^{2\lambda}\le n^\lambda\tag{7}
$$
and each of these inequalities is sharp.
Applying $(7)$ to your specific case yields
$$
\sum_{i=1}^3x_i^2=3\implies6\le\sum_{i=1}^3x_i^4+\sum_{i=1}^3x_i^6\le36\tag{8}
$$
