Prove that: $\sqrt[3]{a_1^3+ a_2^3 +\cdots+a_n^3} \le \sqrt{a_1^2 + a_2^2 +\cdots+a_n^2}$

Question

Let $a_1, a_2, \ldots, a_n \in \mathbb{R}$. Prove that the following inequality holds: $$\sqrt[3]{a_1^3+ a_2^3 +\cdots+a_n^3} \le \sqrt{a_1^2 + a_2^2 +\cdots+a_n^2}$$

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I first tried to restrict the inequality to $\mathbb{R}^{+}$ and that's fairly easy to do since:

$$\sqrt[3]{a_1^3+ a_2^3 +\cdots+a_n^3} \le \sqrt[3]{|a_1^3| + |a_2^3| +\cdots+ |a_n^3|}$$

and

$$\sqrt{a_1^2 + a_2^2 +\cdots+a_n^2} = \sqrt{|a_1^2| + |a_2^2| +\cdots+ |a_n^2|}$$

I tried using the Jensen's Inequality on $x^{\frac 32}$ and $x^{\frac 23}$, but the inequality goes in the opposite direction. Also I tried to divide with $n^{\frac12}$ and $n^{\frac13}$ and use the power mean inequality, but again the inequality goes in the opposite direction.

Anyway, I managed to solve it by raising the inequality to sixth power and we get:

$$\sum a_1^6 + 3\sum a_1^4a_2^2 + 6\sum a_1^2a_2^2a_3^2 \ge \sum a_1^6 + 2\sum a_1^3a_2^3$$

Which is true since:

$$a_1^4 a_2^2 + a_1^2 a_2^4 \ge 2a_1^3 a_2^3$$

and all terms on the LHS are positive. But this prove looks kind of messy. Is there any better proof?

Also it's seems that we can extend this inequality to:

$$k_1 \ge k_2 \implies \sqrt[k_1]{a_1^{k_1}+ a_2^{k_1} +\cdots+a_n^{k_1}} \ge \sqrt[k_2]{a_1^{k_2} + a_2^{k_2} +\cdots+a_n^{k_2}}$$ But can we actually do that?

Answer 1

This inequality follows easily by induction from the following fact (with $p=2/3$):

$(x+y)^p \leq x^p + y^p$ for all $x,y \geq 0$ and $0 < p < 1$.

To prove this, fix $y$, and remark that the function $x \mapsto (x+y)^p-x^p$ is non-increasing, hence is bounded from above by its value at $x=0$.

Answer 2

Since, $\sqrt[3]{\sum\limits_{i=1}^na_i^3}\leq\sqrt[3]{\sum\limits_{i=1}^n|a_i|^3}$ and the right side does not depend on changing sings of our variables, it's enough to prove our inequality for non-negatives $a_i$.

Let $f(x)=x^{\frac{2}{3}}$, $a_i^3=x_i$ and $x_1\geq x_2\geq...\geq x_n$.

Thus, we need to prove that $$\left(\sum_{i=1}^nx_i\right)^{\frac{1}{3}}\leq\left(\sum_{i=1}^nx_i^{\frac{2}{3}}\right)^{\frac{1}{2}}$$ or $$\sum_{i=1}^nf\left(x_i\right)\geq f\left(\sum_{i=1}^nx_i\right)$$ or $$\sum_{i=1}^nf\left(x_i\right)\geq f\left(\sum_{i=1}^nx_i\right)+f(0)+...+f(0),$$ which is true by Karamata because $f$ is a concave function and $$\left(\sum_{i=1}^nx_i,0,...,0\right)\succ(x_1,x_2,...,x_n).$$

Answer 3

Yeah wlg assume $a_i\ge 0$.

Also note that if we multiply all the $a_i$'s with any constant $k$, the inequality is not affected. This will be useful later.

Let $S_p=\sum a_i^p$.

Differentiate w.r.t. $p$. $$\begin{array}{rcl} \frac{d}{dp}S_p^{1/p}&=&S_p^{1/p}\ln(S_p)\cdot \frac{-1}{p^2}+\frac{1}{p}S_p^{1/p-1}\sum a_i^p\ln(a_i)\\ &=&\frac{S_p^{1/p}}{p}\left(-\frac{\ln(S_p)}{p}+\frac{\sum a_i^p\ln(a_i)}{S_p}\right)\\ &=&\frac{S_p^{1/p}}{p}\left(\frac{\sum a_i^p\ln(a_i)}{S_p}-\frac{\ln(S_p)}{p}\right)\\ &=&\frac{S_p^{1/p}}{p\,S_p}\left(\sum a_i^p\ln(a_i)-\frac{S_p\ln(S_p)}{p}\right)\\ &=&\frac{S_p^{1/p}}{p\,S_p}\left(\sum a_i^p\ln(a_i)-\frac{\left(\sum{a_i^p}\right)\ln\left(\sum{a_i^p}\right)}{p}\right) \end{array}$$

Now choose any constant $p_0>1$. If $p=p_0$, we can multiply all $a_i$'s with $k=\frac{1}{\sum a_i^{p_0}}$ and work with the new $a_i$'s.

Since $p_0>1$, and $\sum a_i^{p_0}=1$, each $a_i<=1$. So $\ln a_i\le0$, hence $\sum a_i^{p_0} \ln a_i\le0$.

What all this means is for any $p>1$, any increase in $p$ will decrease $\sqrt[p]{S_p}$. So $\sqrt[3]{\sum a_i^3}=\sqrt{\sum a_i^2}$.