# Cauchy-Schwarz inequality for $a_1^4 + a_2^4 + \cdots + a_n^4 \geqslant n$

Let $$a_1+a_2,...,a_n \in \mathbb{R}.$$ Show that if $$a_1+a_2+...+a_n=n$$, then $$a_1^4+a_2^4+...+a_n^4 \geqslant n.$$

The proposed solution for this was the following:

Using the Cauchy-Schwarz inequality twice we get

$$a_1^4+a_2^4+...+a_n^4 \geqslant \frac{(a_1^2+a_2^2+...+a_n^2)^2}{n} \geqslant \frac{\frac{((a_1+a_2+...+a_n)^2)^2}{n}}{n} = \frac{(\frac{n^2}{n})^2}{n} = n$$

I can see that we can deduce this straight from the definition, but where on earth does the denominator $$n$$ come for $$a_1^4+a_2^4+...+a_n^4 \geqslant \frac{(a_1^2+a_2^2+...+a_n^2)^2}{n}$$.

From Cauchy-Schwarz we can come up with $$a_1^4+a_2^4+...+a_n^4 \geqslant (a_1^2+a_2^2+...+a_n^2)^2$$, but I don't see where the denominator comes from. Could someone enlighten me?

• This is not true: $\sum a_i^4\ge (\sum_{i=1}^n a_i^2)^2$. (The reverse is however true: $\sum a_i^4\le (\sum_{i=1}^n a_i^2)^2$.) If you use C-S you should get $\sum a_i^4\ge \frac1n(\sum_{i=1}^n a_i^2)^2$. Mar 22, 2020 at 14:41
• I would expect a convexity and symmetry argument to imply that the minimum of $\sum a_i^4$ is reached at $(1,\dots,1)$. Mar 22, 2020 at 14:44
• @WETutorialSchool Interesting, the solution was proposed on an official page for the national competition.
– user713999
Mar 22, 2020 at 14:45
• But this result is false. Let n=10, $a_i = 0.1$ Then $n=1$ and $\sum_{i=1} ^{10} (0.1)^4= 0.001$ which is less than $n$ Mar 22, 2020 at 14:52
• @RyderRude the condition is $\sum a_i=n$, not 1. Mar 22, 2020 at 15:00

Also we can use Jensen. Let $$f:\mathbb R \to \mathbb R, x\mapsto x^4$$. Then $$f$$ is convex and thus by Jensen,

$$\frac{a_1^4+\dots+a_n^4}n=\frac{f(a_1)+\dots+f(a_n)}n\geq f\left(\frac{a_1+\dots+a_n}n\right)=\frac{(a_1+\dots+a_n)^4}{n^4}=1.$$

The denominator comes from the following: $$n(a_1^4+...+a_n)^4=(1^2+...+1^2)(a_1^4+_...+a_n^4)\geq(a_1^2+...+a_n^2)^2,$$ which gives $$a_1^4+...+a_n^4\geq\frac{(a_1^2+...+a_n^2)^2}{n}.$$ Another way: $$\sum_{i=1}^n(a_i^4-1)=\sum_{i=1}^n(a_i-1)(a_i^3+a_i^2+a_i+1)=$$ $$=\sum_{i=1}^n((a_i-1)(a_i^3+a_i^2+a_i+1)-4(a_i-1))=\sum_{i=1}^n(a_i-1)^2(a_i^2+2a_i+3)\geq0.$$

• That's an interesting solution. How did you get the idea to use 3? Mar 22, 2020 at 14:49
• @Olivier Bégassat It was typo. It should be $4$. I fixed and added something. Mar 22, 2020 at 14:54
• Ok. My question is more about the intuition to add stuff that sums to zero but produces an interesting factorization of the summand with one factor being $>0$. Is this part of a general method? If so, how does one choose the right amount to add (i.e. 4 vs something else) Mar 22, 2020 at 14:57
• @Olivier Bégassat The intuition in the following. $a_i^3+a_i^2+a_i+1=4$ for $a_i=1$, which gives a factor $a_i-1$ again, which gives a factor $(a_i-1)^2$. This method is named the Tangent Line method. Mar 22, 2020 at 14:59
• Ok, thanks! And I guess then the expectation is that the degree 2 polynomial $P$ with $(a-1)^2\cdot P=(a-1)(a^3+a^2+a+1)-4(a-1)$ to be $\geq 0$ on $\Bbb{R}$. Mar 22, 2020 at 15:01

I always feel that you want to write things very explicitly and lengthily when using Cauchy-Schwarz, or you may forget square roots, or a factor. Here, we use that for any $$(b_i)$$, $$\left | \sum_{i=1}^n b_i \right |= \left | \sum_{i=1}^n b_i \times 1 \right | \leq \sqrt{\sum_{i=1}^n b_i^2} \sqrt{\sum_{i=1}^n 1^2} = \sqrt{\sum_{i=1}^n b_i^2} \sqrt{n},$$ then take the square, apply to $$b_i = a_i^2$$, and then to $$b_i = a_i$$.

Note that if $$\langle \cdot,\cdot\rangle$$ is the scalar product in $$\mathbb{R}^{n}$$ and $$x=(x_{1},...,x_{n}) \in \mathbb{R}^{n}$$ is any vector. We have the following:

Let $$u$$ be the vector full of ones, in other words $$u=(1,1,...,1)$$

Then by C-S we have

$$|\langle x,u\rangle|^{2} \leq \langle u,u\rangle \cdot \langle x,x\rangle$$

It's easy to see that $$\langle u,u\rangle=n$$ and $$\langle x,u \rangle=x_{1}+x_{2}+...+x_{n}$$. Then:

$$\left(x_{1}+x_{2}+...+x_{n}\right)^{2} \leq n \langle x,x\rangle = n (x_{1}^{2}+...+x_{n}^{2})$$

In other words:

$$\frac{\left(x_{1}+x_{2}+...+x_{n}\right)^{2}}{n} \leq x_{1}^{2}+...+x_{n}^{2}$$

When $$x_{i}=a_{i}^{2}$$ the result follows.

• Note: $\langle\cdot, \cdot\rangle$ is a bit nicer for inner products (I am using \langle \rangle) Mar 22, 2020 at 14:56
• @MaximilianJanisch I actually forgot the existence of that, I already change it, thanks. Mar 22, 2020 at 15:01