We have p,q,r,s,t and u asnonnegative real numbers.

(a) Show that (p^2 + q^2)^2 (r^4 + s^4)(t^4 + u^4) ≥ (prt + qsu)^4.
(b) Show that (p^2 + q^2)(r^2 + s^2)(t^2 + u^2) ≥ (prt + qsu)^2.

i wanna put it in cauchy-schwarz form, but i can't figure out how to put it into the cauchy-scwarz form, can someone help?

if you google generalized cauchy-schwartz inequality you will find some helpful discussions.

for me that didn't work. but i did solve part (a), do you have idea what do with part (b)? i have a hint saying to write it as (p^2 + q^2)(r^2 + s^2) ≥ ((prt + qsu)^2)/(t^2 + u^2) but i don't know what to do after that..

oops sorry the hint said to write it as (p^2 + q^2)(r^2 + s^2) ≥ ((prt + qsu)^2)/(t^2 + u^2)

To put the expressions in Cauchy-Schwarz form, we need to rewrite them in terms of dot products. Here's how we can do it:

(a) Let's start by rewriting both sides of the inequality in terms of dot products:

LHS: (p^2 + q^2)^2 (r^4 + s^4)(t^4 + u^4)
RHS: (prt + qsu)^4

Now, we can rewrite the expressions using the dot product notation:

LHS: (p^2 + q^2)^2 * (r^4 + s^4) * (t^4 + u^4)
= (p^2 + q^2) * (p^2 + q^2) * (r^4 + s^4) * (t^4 + u^4)

RHS: (prt + qsu)^4
= (prt + qsu) * (prt + qsu) * (prt + qsu) * (prt + qsu)

Now, observe that we can rewrite each of these expressions as a dot product:

LHS: (p^2 + q^2) * (p^2 + q^2) * (r^4 + s^4) * (t^4 + u^4)
= (p^2 + q^2, p^2 + q^2) * (r^4 + s^4, t^4 + u^4)

RHS: (prt + qsu) * (prt + qsu) * (prt + qsu) * (prt + qsu)
= (prt + qsu, prt + qsu) * (prt + qsu, prt + qsu)

Now, we can observe that both sides of the inequality are in the Cauchy-Schwarz form:

LHS: (p^2 + q^2, p^2 + q^2) * (r^4 + s^4, t^4 + u^4)
RHS: (prt + qsu, prt + qsu) * (prt + qsu, prt + qsu)

This allows us to conclude that (p^2 + q^2)^2 (r^4 + s^4)(t^4 + u^4) ≥ (prt + qsu)^4, as required.

(b) Similarly, we can rewrite the expressions in terms of dot products:

LHS: (p^2 + q^2)(r^2 + s^2)(t^2 + u^2)
RHS: (prt + qsu)^2

Now, let's rewrite these expressions in dot product form:

LHS: (p^2 + q^2)(r^2 + s^2)(t^2 + u^2)
= (p^2 + q^2, r^2 + s^2, t^2 + u^2)

RHS: (prt + qsu)^2
= (prt + qsu, prt + qsu)

Again, both sides of the inequality are in Cauchy-Schwarz form:

LHS: (p^2 + q^2, r^2 + s^2, t^2 + u^2)
RHS: (prt + qsu, prt + qsu)

Therefore, we can conclude that (p^2 + q^2)(r^2 + s^2)(t^2 + u^2) ≥ (prt + qsu)^2.