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!

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.
$$\sum_{k=1}^n(a-k)^3\leq\sum_{k=1}^{a-1}k^3=\frac{a^2(a-1)^2}{4}=\frac{1}{4}(a^4-2a^3+a^2).$$
