Xjyuk

Can you help me prove the following inequality:
$$
(sum_k=1^na_kb_kc_k)^2 leq sum_k=1^na_k^2sum_k=1^nb_k^2sum_k=1^nc_k^2
$$
where $a's,b's,c's in mathrmR$

I tried to use Cauchy's inequality to prove this but got stuck.

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 $(sum limits_k=1^n a_kb_kc_k)^2leq sum limits_k=1^n a_k^2 sum limits_k=1^n (b_k c_k)^2 leq sum limits_k=1^n a_k^2 sum limits_k=1^n b_k^2 sum limits_k=1^n c_k^2$.

You can apply Cauchy's inequality twice. It may be more convenient to use vector notation. Let $d_k = a_kb_k$.
$$||ccdot d||^2 leq ||c||^2||d||^2 = ||c||^2||a cdot b||^2 leq ||c||^2||a||^2||b||^2.$$

So, the inequality should also work over $mathbbC$.

$(sum_k=1^na_kb_kc_k)^2le sum_k=1^na_k^2sum_k=1^n(b_kc_k)^2le sum_k=1^na_k^2sum_k=1^nb_k^2sum_k=1^nc_k^2$.

The last inequality can be seen with $sum_k=1^nb_k^2sum_k=1^nc_k^2=sum_k=1^nsum_j=1^n(b_kc_j)^2ge sum_k=1^n(b_kc_k)^2$.

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
$$
sum_k=1^n x_k^2 y_k^2
leq
left(sum_k=1^n x_k y_kright)^2,
$$
obvious, since sum on the right contains everything on the left plus something more.

Lemma L2. Consider two vectors $vecx$ and $vecy$ that have components $x_1,dots,x_n$ and $y_1,dots,y_n$ and following scalar products
$$
left(vecx cdot vecyright)^2 =
left(sum_k=1^n x_k y_kright)^2
;quad
left(vecx cdot vecxright) left(vecy cdot vecyright) =
left(sum_k=1^n x_k^2right) left(sum_k=1^n y_k^2right).
$$
But we know that
$$
left(vecx cdot vecyright)^2 = |vecx|^2 |vecy|^2 cos^2(phi)
= left(vecx cdot vecxright) left(vecy cdot vecyright) cos^2(phi),
$$
where $phi$ is an angle between $vecx$ and $vecy$.
Since $cos^2(phi) leq 1$ we can state
$$
left(sum_k=1^n x_k y_kright)^2 leq
left(sum_k=1^n x_k^2right) left(sum_k=1^n y_k^2right).
$$

Now we are equipped to do the proof
$$
left(sum_k=1^n a_k b_k c_kright)^2
stackreltextL2leq
left(sum_k=1^n a_k^2right) left(sum_k=1^n (b_k c_k)^2right)
stackreltextL1leq
left(sum_k=1^n a_k^2right) left(sum_k=1^n b_k c_kright)^2
stackreltextL2leq
left(sum_k=1^n a_k^2right) left(sum_k=1^n b_k^2right) left(sum_k=1^n c_k^2right)
$$
QED

Since the RHS is non-negative and is not changed after substitution $a_krightarrow -a_k$ and a similar for another variables, it's enough to assume that for any $k$, $a_kgeq0$, $b_kgeq0$ and $c_kgeq0$.

Now, let $a_kb_kc_k=x_k.$

Thus, by Holder
$$sum_k=1^na_k^2sum_k=1^nb_k^2sum_k=1^nc_k^2geqleft(sum_k=1^nsqrt[3]a_k^2b_k^2c_k^2right)^3$$ and it's enough to prove that
$$left(sum_k=1^nx_k^frac23right)^3geqleft(sum_k=1^nx_kright)^2$$ or
$$sum_k=1^nx_k^frac23geqleft(sum_k=1^nx_kright)^frac23.$$
Now, let $f(x)=x^frac23.$

Thus, $f$ is a concave function.

Also, let $x_1geq x_2geq...geq x_n.$

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.