Show that for any non negative real numbers $x_1,x_2,\cdots x_n,$
$$\frac{{x_1}^2+{x_2}^2+\cdots+{x_n}^2}{n}x_1x_2\cdots x_n\le\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)^{n+2}$$
My work:
Let$$S(n)=\frac{{x_1}^2+{x_2}^2+\cdots+{x_n}^2}{n}x_1x_2\cdots x_n\le\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)^{n+2}$$
By theorem of triviality, if any of $x_i$'s are $0$ the inequality is certainly true. So assume all numbers are $\gt0$
$S(1)$ says ${x_1}^3\le {x_1}^3$ which is certainly true.
$S(2)$ says $({x_1}^2+{x_2}^2)x_1x_2\le\frac18(x_1+x_2)^4$ which reduces to $0\le(x_1-x_2)^4$ which is certainly true.
Assume $S(k)$ is true. Now we just needs to prove that $S(k+1)$ is true. But I'm having a hard time in doing that.
Any help is greatly appreciated. Or is there any better method than induction$?$
Best Answer
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.