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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.

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5 Answers 5

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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 $(\sum \limits_{k=1}^n a_kb_kc_k)^2\leq \sum \limits_{k=1}^n {a_k}^2 \sum \limits_{k=1}^n (b_k c_k)^2 \leq \sum \limits_{k=1}^n {a_k}^2 \sum \limits_{k=1}^n b_k^2 \sum \limits_{k=1}^n c_k^2$.

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    $\begingroup$ fantastic............................................ $\endgroup$
    – Guy Fsone
    Commented Aug 5, 2019 at 17:20
  • $\begingroup$ Can you show when the equality holds? $\endgroup$ Commented Aug 6, 2019 at 14:34
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$(\sum_{k=1}^na_kb_kc_k)^2\le \sum_{k=1}^na_k^2\sum_{k=1}^n(b_kc_k)^2\le \sum_{k=1}^na_k^2\sum_{k=1}^nb_k^2\sum_{k=1}^nc_k^2$.

The last inequality can be seen with $\sum_{k=1}^nb_k^2\sum_{k=1}^nc_k^2=\sum_{k=1}^n\sum_{j=1}^n(b_kc_j)^2\ge \sum_{k=1}^n(b_kc_k)^2$.

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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.

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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 $\mathbb{C}$.

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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

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