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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

RE60K
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    "It is intuitively clear that the upper bound will be n when all variables are equal to 1." Actually this case yields the lower bound. The upper bound is $n^\lambda$, rather. – Did Sep 02 '14 at 08:56
  • Expand $(\sum x_i^2)^{\lambda}$. You get $\sum x_i^{2\lambda}$ plus a lot of other terms which are all positive. – almagest Sep 02 '14 at 09:07
  • $x^4(1+x^2)$ is a convex function, hence $\sum x_i^4(1+x_i^2)$ attains its maximum on the boundary - on the other hand, the listed inequalities have stationary points inside the domain. – Jack D'Aurizio Sep 02 '14 at 09:29
  • You have significantly changed the question. Furthermore, in keeping with your previous question, did you mean $$\sum_{i=1}^3x_i^2=3$$ on the right? – robjohn Sep 02 '14 at 10:13
  • @robjohn edited – RE60K Sep 02 '14 at 10:18
  • @Aditya: I have edited the title and question to match. If that is not what you intended, revert my edit. – robjohn Sep 02 '14 at 10:24
  • @robjohn nope it's coreect. – RE60K Sep 02 '14 at 10:25

1 Answers1

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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 $$ \sum_{i=1}^3x_i^2=3\implies6\le\sum_{i=1}^3x_i^4+\sum_{i=1}^3x_i^6\le36\tag{8} $$

robjohn
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