$x_1x_2+x_2x_3+x_3x_4+x_4x_5=(x_1+x_3+x_5)(x_2+x_4) - (x_2x_5+x_1x_4) $
Now we try to find maximum value of $(x_1+x_3+x_5)(x_2+x_4)$ when $(x_1+x_3+x_5)+(x_2+x_4)=5$
And try to minimize the value of $(x_2x_5+x_1x_4) $.
Take, $a=(x_1+x_3+x_5)$ and $b=(x_2+x_4)$
By , A.M. $\ge $ G.M. $\implies$ $\sqrt(ab) \le \frac{a+b}{2} \implies (ab) \le (\frac{5}{2})^2 $
So, max value of $(x_1+x_3+x_5)(x_2+x_4)$ is $(\frac{5}{2})^2 $
And , clearly, minimum value of $(x_2x_5+x_1x_4) $ is $0$.
So , max value of $x_1x_2+x_2x_3+x_3x_4+x_4x_5$ is $(\frac{5}{2})^2 $
Let $k\ge2$ be fixed and suppose that $S(k)$ holds true.
$$S(k):({x_1}^2+{x_2}^2+\cdots+{x_k}^2)x_1x_2\cdots x_k\le(k)\left(\frac{x_1+x_2+\cdots+x_k}{k}\right)^{k+2}$$ It remains to prove:
$$S(k+1):({x_1}^2+{x_2}^2+\cdots+{x_k}^2+x^2)x_1x_2\cdots x_k\cdot x\le(k+1)\left(\frac{x_1+x_2+\cdots+x_k+x}{k+1}\right)^{k+3}$$ Using the fact that $x=x_{k+1}$
Now, Put $A=\frac{x_1+x_2+\cdots+x_k}{k}$ and $P=x_1x_2\cdots x_k$
$\implies$ $$S(k):({x_1}^2+{x_2}^2+\cdots+{x_k}^2)P\le kA^{k+2}$$
And it remains to prove that $$S(k+1):({x_1}^2+{x_2}^2+\cdots+{x_k}^2+x^2)Px\le (k+1)\left(\frac{kA+x}{k+1}\right)^{k+3}$$ The left hand side of $S(k+1)$ is
$$({x_1}^2+{x_2}^2+\cdots+{x_k}^2)Px+Px^3\le kA^{k+2}x+Px^3$$ The above inequality came by using the fact of $S(k)$
So to prove $S(k+1)$, it suffices to prove that
$$kA^{k+2}x+Px^3\le (k+1)\left(\frac{kA+x}{k+1}\right)^{k+3}$$
By $AM\ge GM$, $P\le A^k$, so it suffices to prove that
$$kA^{k+2}x+A^kx^3\le (k+1)\left(\frac{kA+x}{k+1}\right)^{k+3}$$
Now restrict to the situation where the sum $x_1+x_2+\cdots+x_k+x$ is held constant, and prove
the result with this added constraint. The general result then follows immediately; observe that for any constant $c$, the statement $S(n)$ holds for $x_1,\cdots,x_n$ if and only if it holds for $cx_1,\cdots,cx_n$ (the factor $c^{n+2}$ appears on each side). So, consider only those $(x_1,\cdots,x_k,x)\in\mathbb{R}^{k+1}$ for which $x_1+x_2+\cdots+x_k+x=k+1$,that is,
$$kA+x=k+1$$
So, to prove $S(k + 1)$, it suffices to show $$kA^{k+2}x+A^kx^3\le k+1$$
The left hand of the above inequality is a function of $A$ (and $x = k + 1 − kA$, also a function of $A$), and so this expression is maximized using calculus:
$$\frac{d}{dA}=[kA^{k+2}x+A^kx^3]=k(k+2)A^{k+1}x+kA^{k+2}\frac{dx}{dA}+kA^{k-1}x^3+A^k\cdot3x^2\frac{dx}{dA}$$
$$=k(k+2)A^{k+1}x+kA^{k-1}x^3-k(kA^{k+2}+A^k\cdot3x^2)$$
Putting $A = tx$, this expression becomes (after a bit of algebra)
$$(1-t)(kt^2-2t+1)kt^{k-1}x^{k+2}$$
Since $k\ge2$, the above has roots at only $t = 0$ and $t = 1$, and so the derivative is positive for $0 < t < 1$ and negative for $t > 1$. Thus, $kA^{k+2}x+A^kx^3$ achieves a maximum when $t = 1$, that is, when $A = x = 1$. Hence, $$kA^{k+2}x+A^kx^3\le k+1$$ and so $S(k + 1)$ follows, completing the inductive step.
Thus, by mathematical induction, for all $n ≥ 1$, the statement $S(n)$ is true.
Best Answer
The idea is the same: $$ x_1x_2 + x_2x_3 + \cdots + x_{2020}x_1 \leq \frac{x_1^2 + x_2^2}{2} + \frac{x_2^2 + x_3^2}{2} + \cdots + \frac{x_{2020}^2 + x_1^2}{2} = x_1^2 + x_2^2 + \cdots + x_{2020}^2 $$