Minimizing over partitions $f(\lambda) = \sum \limits_{i = 1}^N |\lambda_i|^4/(\sum \limits_{i = 1}^N |\lambda_i|^2)^2$ I'm trying to characterize the behavior of the the quantity:
$$A = \frac{\sum \limits_{i = 1}^N x_i^4}{(\sum \limits_{i = 1}^N x_i^2)^2},$$
subject to the constraints that 
$$ \sum \limits_{i = 1}^N x_i = N, \ x_i \ge 0 \ \forall \ i$$

(I.e., we are working with all integer partitions of size $N$). 
My hypothesis is that if there exists $x_i \ge N/2$, then $A \ge \frac{x_i^4 + (N-x_i)^4}{(x_i^2 + (N-x_i)^2)^2}$ (which I've verified experimentally for N up to 40). But I'm having trouble proving this hypothesis. 

The closest I've come is to make use of the fact that $\sum \limits_{i = 1}^N x_i^2 \le N^2$, which gives an upper bound for the denominator, but the corresponding inequality for fourth powers in the numerator doesn't seem helpful, since we're trying to find a lower bound. 

I'd greatly appreciate any advice. Please ask for any clarifications if anything is unclear. 

Edit:
I've decided this question is better phrased explicitly in terms of partitions. So let $$f(\lambda) = \frac{\sum \limits_{i = 1}^N |\lambda_i|^4}{(\sum \limits_{i = 1}^N |\lambda_i|^2)^2},$$
be a function over partitions $\lambda$ of size $N$ and $|\lambda_i|$ is the size of the $i$-th part. The hypothesis now is that in the domain of all partitions $\lambda'$ in which there exists $\lambda_i$ such that $|\lambda_i| \ge N/2$, $f(\lambda')$ is minimized by the 2-element partition $&lt;\lambda_i, \lambda_j&gt;$ (in which $|\lambda_j| = N - |\lambda_i|$).
You can assume that $x_N\ge N/2$ and so you want to prove that
$$
 \frac{\sum \limits_{i = 1}^N x_i^4}{\left(\sum \limits_{i = 1}^N x_i^2\right)^2}\ge \frac{x_N^4 + (N-x_N)^4}{(x_N^2 + (N-x_N)^2)^2}
$$
Set $\lambda:=N-x_N$, and assume $\lambda&gt;0$ (else the inequality is trivially true), and set also 
$$
x:= \frac{x_N}{\lambda},\quad\text{and}\quad y_i:=\frac{x_i}{\lambda},\quad\text{for $i=1,\dots, N-1$}.
$$
The inequality now reads:
$$
\frac{x^4+\sum \limits_{i = 1}^{N-1} y_i^4}{\left(x^2+\sum \limits_{i = 1}^{N-1} y_i^2\right)^2}
\ge \frac{x^4 + 1}{(x^2 + 1)^2},
$$
and you want to prove this inequality under the conditions $x\ge 1$ (corresponding to $x_N\ge N/2$) and $S_1:=\sum_{i=1}^{N-1}y_i=1$,
(corresponding to $\sum_{i=1}^N x_i=N$). Set $S_2:=\sum_{i=1}^{N-1}y_i^2$ and $S_4:=\sum_{i=1}^{N-1}y_i^4$. Then you want to prove that
the inequality 
$$
(S_4+x^4)(x^2+1)^2\ge (x^4+1)(S_2+x^2)^2
$$
holds, or equivalently, that for the function
$$
f(x):=S_4 x^4+2 S_4 x^2+S_4+2x^6-2S_2 x^6-2x^2S_2-S_2^2-S_2^2x^4
$$
we have $f(x)\ge 0$ for $x\ge 1$.
Below we will prove the following inequality:
$$
(MI) \qquad\qquad\qquad 1+2S_4\ge 2 S_2+S_2^2. 
$$
From this inequality it follows that $f(1)=4S_4+2-4S_2-2S_2^2\ge 0$.
Hence it suffices to prove that $f'(x)\ge 0$ for all $x\ge 1$. 
But 
$$
f'(x)=4S_4 x^3+4 S_4 x+12 x^5-12 S_2 x^5 -4 S_2 x -4 S_2^2 x^3,
$$
and so $f'(x)=2x g(x)$ with
$$
g(x):=2 S_4 x^2+2S_4+6 x^4-6 S_2 x^4-2S_2-2S_2^2 x^2
$$ 
and we have to prove that $g(x)\ge 0$ for $x\ge 1$.
From $(MI)$ we obtain 
$$
2S_4 x^2-S_2^2 x^2\ge 2x^2 S_2-x^2\quad\text{and}\quad 
2S_4-2S_2\ge S_2^2-1,
$$
which implies
$$
g(x)\ge 2 S_2 x^2-x^2+S_2^2-1+6x^4(1-S_2).
$$
Now clearly $S_2\le 1$ and so $6x^4(1-S_2)\ge 6x^2(1-S_2)$. We arrive at
$$
g(x)\ge S_2^2+(x^2-1)+4x^2(1-S_2)\ge 0,
$$
which concludes the proof.

Finally we prove 
$$
(MI) \qquad\qquad\qquad 1+2S_4\ge 2 S_2+S_2^2.
$$
We will prove a slightly more general result:
Take $x_1,\dots,x_n$ with $x_i\ge 0$ and set
$$
S_1:=\sum_{i=1}^n x_i,\quad S_2:=\sum_{i=1}^n x_i^2\quad\text{and}\quad S_4:=\sum_{i=1}^n x_i^4.
$$
Then we have
$$
S_1^4+2S_4\ge 2S_2 S_1^2+S_2^2.
$$
This inequality, in the case $S_1=1$, gives $(MI)$.

Let's prove the inequality $
S_1^4+2S_4\ge 2S_2 S_1^2+S_2^2
$:

Adding $S_2^2$ and subtracting $2S_4$ and $2 S_2 S_1^2$ it is equivalent to
$$
S_1^4-2S_2 S_1^2+S_2^2\ge S_2^2-2S_4+S_2^2,
$$
hence we have to prove
$$
(S_1^2-S_2)^2\ge 2(S_2^2-S_4).
$$
Now we compute
$$
S_1^2-S_2=\left(\sum x_i\right)\left(\sum x_j\right)-\sum x_i^2=\sum_{i\ne j}x_i x_j
$$
and
$$
S_2^2-S_4=\left(\sum x_i^2\right)\left(\sum x_j^2\right)-\sum x_i^4=\sum_{i\ne j}x_i^2 x_j^2.
$$
Subtracting $\sum_{i\ne j}x_i^2 x_j^2$ from both sides of the inequality
$$
\left(\sum_{i\ne j}x_i x_j\right)^2\ge 2\sum_{i\ne j}x_i^2 x_j^2,
$$
we see that the inequality is equivalent to
$$
\sum_{\begin{array}{c} (i,j)\ne (k,l)\\ i\ne j\\ k\ne l\end{array}}x_i x_j x_k x_l\ge \sum_{i\ne j}x_i^2 x_j^2.
$$
But we have
$$
\sum_{\begin{array}{c} (i,j)\ne (k,l)\\ i\ne j\\ k\ne l\end{array}}x_i x_j x_k x_l=
\sum_{\begin{array}{c} (i,j)\ne (k,l)\\ i\ne j\\ k\ne l \\ (i,j)=(l,k)\end{array}}x_i x_j x_k x_l+
\sum_{\begin{array}{c} (i,j)\ne (k,l)\\ i\ne j\\ k\ne l \\ (i,j)\ne (l,k)\end{array}}x_i x_j x_k x_l
$$
hence
$$
\sum_{\begin{array}{c} (i,j)\ne (k,l)\\ i\ne j\\ k\ne l\end{array}}x_i x_j x_k x_l\ge
\sum_{\begin{array}{c} (i,j)\ne (k,l)\\ i\ne j\\ k\ne l \\ (i,j)=(l,k)\end{array}}x_i x_j x_k x_l=\sum_{i\ne j}x_i^2 x_j^2,
$$
which concludes the proof of $(MI)$.
The trick here is to consider the following "switching operation":
$$\sigma_{ij}(\lambda_1,\ldots,\lambda_N) = (\lambda_1,\ldots,\lambda_i+1,\ldots,\lambda_j-1,\ldots,\lambda_N).$$
Now the question becomes: what can we say about $f(\sigma_{ij}(\lambda))$?

Assume without loss of generality that $\lambda_j\ge1$, and that all $\lambda_k\ge0$. A short computation shows:
$$f(\sigma_{ij}(\lambda)) = \frac{\sum_{k=1}^N\lambda_k^4 + 4(\lambda_i^3-\lambda_j^3) + 6(\lambda_i^2+\lambda_j^2) + 4(\lambda_i-\lambda_j) + 2}{\left(\sum_{\ell=1}^N\lambda_\ell^2+2(\lambda_i - \lambda_j) + 2\right)^2}$$
Now write $\lambda_j = \lambda_i+k$ with $k$ an integer. Then the above expression becomes
$$f(\sigma_{ij}(\lambda)) = \frac{\sum_{k=1}^N\lambda_k^4 + 2\lambda_i^2+k(4+2\lambda_i-3\lambda_i^2)+k^2(1-3\lambda_i^2)-k^3}{\left(\sum_{\ell=1}^N\lambda_\ell^2+2(k+1)\right)^2}$$
Now we are left with the tedious work of checking by cases ($k&gt;0$, $k&lt;0$ etc.) when this decreases with respect to $f(\lambda)$. This allows you to minimize the quantity.