Upper Bound for a Sum Can you help me prove the following inequality:
$$
(\sum_{k=1}^na_kb_kc_k)^2 \leq \sum_{k=1}^na_k^2\sum_{k=1}^nb_k^2\sum_{k=1}^nc_k^2
$$
where $a's,b's,c's \in \mathrm{R}$
I tried to use Cauchy's inequality to prove this but got stuck.
 A: For all $k$ between $1$ and $n$, we have that ${c_k}^2 \leq \sum \limits_{i=1}^n {c_i}^2$, therefore you get that $\sum \limits_{k=1}^n {b_k}^2 {c_k}^2 \leq \sum \limits_{k=1}^n {b_k}^2 \sum \limits_{k=1}^n {c_k}^2$, since all the ${b_k}^2$ are non negative. Now by Cauchy's inequality $(\sum \limits_{k=1}^n a_kb_kc_k)^2\leq \sum \limits_{k=1}^n {a_k}^2 \sum \limits_{k=1}^n (b_k c_k)^2 \leq \sum \limits_{k=1}^n {a_k}^2 \sum \limits_{k=1}^n b_k^2 \sum \limits_{k=1}^n c_k^2$.
A: $(\sum_{k=1}^na_kb_kc_k)^2\le \sum_{k=1}^na_k^2\sum_{k=1}^n(b_kc_k)^2\le  \sum_{k=1}^na_k^2\sum_{k=1}^nb_k^2\sum_{k=1}^nc_k^2$.
The last inequality can be seen with $\sum_{k=1}^nb_k^2\sum_{k=1}^nc_k^2=\sum_{k=1}^n\sum_{j=1}^n(b_kc_j)^2\ge \sum_{k=1}^n(b_kc_k)^2$.
A: Since the RHS is non-negative and is not changed after substitution $a_k\rightarrow -a_k$ and a similar for another variables, it's enough to assume that for any $k$, $a_k\geq0$, $b_k\geq0$ and $c_k\geq0$.
Now, let $a_kb_kc_k=x_k.$
Thus, by Holder
$$\sum_{k=1}^na_k^2\sum_{k=1}^nb_k^2\sum_{k=1}^nc_k^2\geq\left(\sum_{k=1}^n\sqrt[3]{a_k^2b_k^2c_k^2}\right)^3$$ and it's enough to prove that
$$\left(\sum_{k=1}^nx_k^{\frac{2}{3}}\right)^3\geq\left(\sum_{k=1}^nx_k\right)^2$$ or
$$\sum_{k=1}^nx_k^{\frac{2}{3}}\geq\left(\sum_{k=1}^nx_k\right)^{\frac{2}{3}}.$$
Now, let $f(x)=x^{\frac{2}{3}}.$
Thus, $f$ is a concave function.
Also, let $x_1\geq x_2\geq...\geq x_n.$
Thus, $$(x_1+x_2+...+x_n,0,...,0)\succ(x_1,x_2,...,x_n)$$ and by Karamata we obtain:
$$f(x_1)+f(x_2)+...+f(x_k)\geq f(x_1+x_2+...+x_n)+f(0)+...+f(0),$$ which ends a proof. 
A: You can apply Cauchy's inequality twice.  It may be more convenient to use vector notation.  Let $d_k = a_kb_k$.
$$||c\cdot d||^2 \leq ||c||^2||d||^2 = ||c||^2||a \cdot b||^2 \leq ||c||^2||a||^2||b||^2.$$
So, the inequality should also work over $\mathbb{C}$.
A: I'd like to add one more version.
First, notice that it is enough to prove the inequality for positive numbers -- left side can become only smaller if any negative number is present, while the right side does not change.
Without loss of generality, I will consider everything positive from now on.
Now let's prove $2$ lemmas.
Lemma L1
$$
\sum_{k=1}^n x_k^2 y_k^2
\leq
\left(\sum_{k=1}^n x_k y_k\right)^2,
$$
obvious, since sum on the right contains everything on the left plus something more.
Lemma L2. Consider two vectors $\vec{x}$ and $\vec{y}$ that have components $x_1,\dots,x_n$ and $y_1,\dots,y_n$ and following scalar products
$$
\left(\vec{x} \cdot \vec{y}\right)^2 =
\left(\sum_{k=1}^n x_k y_k\right)^2
;\quad
\left(\vec{x} \cdot \vec{x}\right) \left(\vec{y} \cdot \vec{y}\right) =
\left(\sum_{k=1}^n x_k^2\right) \left(\sum_{k=1}^n y_k^2\right).
$$
But we know that 
$$
\left(\vec{x} \cdot \vec{y}\right)^2 = |\vec{x}|^2 |\vec{y}|^2 \cos^2(\phi)
= \left(\vec{x} \cdot \vec{x}\right) \left(\vec{y} \cdot \vec{y}\right) \cos^2(\phi),
$$
where $\phi$ is an angle between $\vec{x}$ and $\vec{y}$.
Since $\cos^2(\phi) \leq 1$ we can state
$$
\left(\sum_{k=1}^n x_k y_k\right)^2 \leq
\left(\sum_{k=1}^n x_k^2\right) \left(\sum_{k=1}^n y_k^2\right).
$$
Now we are equipped to do the proof
$$
\left(\sum_{k=1}^n a_k b_k c_k\right)^2 
\stackrel{\text{L2}}{\leq}
\left(\sum_{k=1}^n a_k^2\right) \left(\sum_{k=1}^n (b_k c_k)^2\right)
\stackrel{\text{L1}}{\leq}
\left(\sum_{k=1}^n a_k^2\right) \left(\sum_{k=1}^n b_k c_k\right)^2
\stackrel{\text{L2}}{\leq}
\left(\sum_{k=1}^n a_k^2\right) \left(\sum_{k=1}^n b_k^2\right) \left(\sum_{k=1}^n c_k^2\right) 
$$
QED
