# Proving an inequality using the Cauchy-Schwarz inequality

Consider four real numbers $a_1, a_2, a_3, a_4$ such that $\sum a_i^3 = 10$. Prove that

$$\sum a_i^4 \geq \sqrt{2500}$$

Applying the Cauchy Schwarz inequality with $a_i^2$ and $a_i$, we get

$$\left(\sum a_i^3\right)^2 \leq \left(\sum a_i^4\right)\left(\sum a_i^2\right)$$

Again, applying the Cauchy Schwarz inequality with $a_i^2$ and $1$, we get:

$$\left(\sum a_i^2\right)^2 \leq 4\left(\sum a_i^4\right)$$

Substituting this into the first inequality, we get:

$$\left(\sum a_i^3\right)^2 \leq 4\left(\sum a_i^4\right)^2$$

Taking the square root of both sides,

$$\left(\sum a_i^3\right) \leq 2\left(\sum a_i^4\right)$$

$$\implies 5 \leq \sum a_i^4$$

But, $\sqrt{2500} = 13.5720881$, which is more than $2$ times $5$. How do I prove the required statement?

• Hmm, I'm not sure how to get it yet, but notice that if any of them are negative then they contribute negatively to the sum of the cubes, but positively to the sum of the fourths. This seems relevant somehow. Also, does it need to be only Cauchy-Schwarz? Because this looks a lot like AM-GM to me. In particular $\sqrt{2500} = (10^4/4)^{1/3}$ Oct 25, 2013 at 9:41

Using the following relations from Cauchy-Schwarz (which has the advantage of being applicable for all reals): $$\left(\sum a_i^4\right)\left(\sum 1\right)\ge \left(\sum a_i^2\right)^2 \tag{1}$$ $$\left(\sum a_i^4\right)\left(\sum a_i^2\right)\ge \left(\sum a_i^3\right)^2 = 100\tag{2}$$

Noting both LHS and RHS are positive, square both sides of (2) and multiply with (1), then cancel off the positive $(\sum a_i^2)^2$ from both sides to get:
$$4 \left(\sum a_i^4\right)^3 \ge 100^2 \implies \sum a_i^4 \ge \sqrt{2500}$$

I don't think the Cauchy-Schwarz inequality gives a quick proof. If you don't insist on that, the quick proof follows directly from the following inequality: $$\sqrt\frac{\sum a_i^3}4\leq\sqrt\frac{\sum a_i^4}4$$ That is, $$\sum a_i^4\geq4\left(\frac{\sum a_i^3}{4}\right)^{\frac43}=\sqrt{2500}$$

Please let me explain something because some mentioned that power mean inequality holds only for non-negative $a_i$. But I insist that if $M_3(a_i)>0, M_4(a_i)>0$, then we can still apply the inequality.

Let us separate $a_i$ according to whether it's non-negative or negative. We can write $$\sum a_i=\sum a_++\sum a_-\tag{*}$$ then still consider the same inequality \begin{align}\sqrt\frac{\sum a_i^3}4&=\sqrt\frac{\sum a_+^3+\sum a_-^3}4\\ &<\sqrt\frac{\sum a_+^3+\sum(-a_-)^3}4\\ &\leq\sqrt\frac{\sum a_+^4+\sum(-a_-)^4}4\\ &=\sqrt\frac{\sum a_+^4+\sum a_-^4}4=\sqrt\frac{\sum a_i^4}4\end{align} The first and the last line is a simple expansion and collection using (*). The second line follows that negative is always less than positive. And the third line is the direct result of power mean inequality.

Note the second line is strictly less than, which implies if "=" attains, all $a_i$ must be positive.

• Note that you can't apply AM-GM directly since some of the $a_i$ could be negative. Oct 25, 2013 at 15:10
• @Macavity In my opinion, power mean inequality is stated for positive numbers $a_i$ just because in some cases power mean gives opposite sign or even worse, nonsense expression. For example, if all $a_i<0$, obviously $M_2(a_i)\geq M_3(a_i)$ since $M_2>0$ and $M_3<0$. But in this particular case, $0<M_3(a_i)\leq M_4(a_i)$ always holds. Oct 25, 2013 at 15:56
• (not sure what the deleted comment is) So all you have shown is the case when all the $a_i$ are positive. I don't see an easy way of dealing with negative $a_i$, apart from trying to individually smooth the terms. Oct 25, 2013 at 17:53
• @CalvinLin No, I mean what I've shown is as general as for real because either side of power mean inequality is well defined and meanwhile positive. Oct 26, 2013 at 0:33
• @ShuchangZhang I think the usual proofs of Power Mean inequality use non-negativeness (e.g. using strict convexity of $x^p$ for $x > 0, p> 1$). So even if the final expression looks well defined for reals, you may have to revisit the proof to show it holds. Oct 26, 2013 at 3:28