Proving inequality: $\sum_{i=1}^n \left(a_i^7+a_i^5\right) \geq 2(\sum_{i=1}^n a_i^3)^2$ 
Let $a_i$ be distinct positive integers; prove that $$(a_1^7+a_2^7+\cdots + a_n^7)+(a_1^5+a_2^5+\cdots +a_n^5)\ge 2(a_1^3+a_2^3+\cdots + a_n^3)^2$$

I tried using some well known inequalities; obviously, since non homogenous and no obvious function, I don't expect either of AGM, Muirhead, CS, Jensen, Karamata, etc. should work, though I might be woefully wrong. After a while of experimentation I realized that this problem would likely be solved by either some tricky manipulations or a very obscure named inequality. Any helps? Thanks!
 A: We can use induction here.
For $n=1$ it's true by AM-GM.
Now, let $$\sum_{k=1}^n(a_k^7+a_k^5)\geq\left(\sum_{k=1}^na_k^3\right)^2.$$
We'll prove that:
$$\sum_{k=1}^{n+1}(a_k^7+a_k^5)\geq\left(\sum_{k=1}^{n+1}a_k^3\right)^2.$$
Indeed, let $a_{n+1}=a=\max\limits_k\{a_k\}$.
Thus, $$\sum_{k=1}^{n+1}(a_k^7+a_k^5)\geq \left(\sum_{k=1}^na_k^3\right)^2+a^7+a^5$$ and it's enough to prove that
$$a^7+a^5\geq2a^6+4a^3\sum_{k=1}^na_k^3$$ or
$$a^4+a^2\geq2a^3+4\sum_{k=1}^na_k^3.$$
Now, since $a_k\leq a-n-1+k,$ it's enough to prove that
$$\sum_{k=1}^n(a-k)^3\leq\frac{1}{4}(a^4-2a^3+a^2)$$ or
$$na^3-\frac{3n(n+1)}{2}a^2+\frac{3n(n+1)(2n+1)}{6}a-\frac{n^2(n+1)^2}{4}\leq\frac{1}{4}(a^4-2a^3+a^2)$$ or
$$(a-n)^2(a-n-1)^2\geq0$$ and we are done!
As David Cheng said, we can prove the last inequality by the following simpler way.
$$\sum_{k=1}^n(a-k)^3\leq\sum_{k=1}^{a-1}k^3=\frac{a^2(a-1)^2}{4}=\frac{1}{4}(a^4-2a^3+a^2).$$
A: One more induction proof
For $n=1$ everything is clear. Suppose that we know that inequality holds for $0<a_1<a_2<\dots<a_n$. We're going to add element $a_{n+1} > a_n$. Denote by $S_n^d = \sum_{i=0}^{n}a_i^d$ sum of $d$-th degrees of first $n$ elements. We know that $$S_n^7 + S_n^5 \ge 2(S_n^3)^2.\tag{1}$$ Adding $a_{n+1}$ gives
$$
S_n^7 + S_n^5 + a_{n+1}^7 + a_{n+1}^5 \ge 2(S_n^3 + a_{n+1}^3)^2
$$
or, using $(1)$,
$$
a_{n+1}^7 + a_{n+1}^5 \ge 4S_n^3a_{n+1}^3 + 2a_{n+1}^6 $$
$$
 \frac{(a_{n+1} - 1)^2}{4a_{n+1}} \ge S_n^3a_{n+1}^{-3}\tag{2}
$$
Now let's take a look at $S_n^3$. Since $a = a_{n+1} > a_n > \dots > a_0 > 0$, we have
$$
S_n^3a_{n+1}^{-3} = \sum_{k=0}^{n}\left(\frac{a_k}{a_{n+1}}\right)^3 < \sum_{k=1}^{n+1}\left(1 - \frac{k}{a}\right)^3 < \sum_{k=1}^{a}\left(1 - \frac{k}{a}\right)^3 = \frac{(a - 1)^2}{4a}
$$
which is exact left hand side of $(2)$.
