Proving inequality: $\sum_{i=1}^n \left(a_i^7+a_i^5\right) \geq 2(\sum_{i=1}^n a_i^3)^2$

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

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 $(2)$.