1. ## Inequality

if $p > q > r > 0$

prove $p^4 + q^4 + r^4 > p^2 q^2 +q^2 r^2 +r^2 p^2$

2. $p > q > 0\; \Rightarrow \;p^2 > pq\; \Rightarrow \;p^4 > p^2 q^2$
Can you use this to complete the problem?

3. Originally Posted by Plato
$p > q > 0\; \Rightarrow \;p^2 > pq\; \Rightarrow \;p^4 > p^2 q^2$
Can you use this to complete the problem?
I get up to $p^4 + q^4 > p^2 q^2 +q^2 r^2
$

but then i keep going around in circles.

4. Use the Cauchy–Schwarz inequality:

$p^2q^2+q^2r^2+r^2p^2\ \leq\ \left[(p^2)^2+(q^2)^2+(r^2)^2\right]^{\frac{1}{2}}\left[(q^2)^2+(r^2)^2+(p^2)^2\right]^{\frac{1}{2}}$

Equality obtains if and only if the vectors $\left(\begin{array}{c}p^2\\q^2\\r^2\end{array}\rig ht)$ and $\left(\begin{array}{c}q^2\\r^2\\p^2\end{array}\rig ht)$ in $\mathbb{R}^3$ are linearly dependent; i.e. if and only if $p^2=q^2=r^2$. Since $p>q>r>0$, this won’t happen and so the inequality is always strict.

5. Originally Posted by bobak
if $p > q > r > 0$

prove $p^4 + q^4 + r^4 > p^2 q^2 +q^2 r^2 +r^2 p^2$
$a^2 + b^2 + c^2 \geq ab+bc+ca$
(By AM-GM, $\Sigma_{a,b,c}(a^2 + b^2) \geq \Sigma_{a,b,c}2ab$)

So substitute $a=p^2,b=q^2,c=r^2$

The above AM-GM holds iff a=b=c.
Or if $p^2=q^2=r^2$

But since p>q>r, this cannot happen. So the inequality is strict. By the way, they need not be positive for AM-GM application here(since we are applying it to squares of those numbers)

6. Originally Posted by Isomorphism
By the way, they need not be positive for AM-GM application here(since we are applying it to squares of those numbers)
They need to be positive for the inequality to be strict. For example, if $p=1,\ q=-1,\ r=1$, the inequality will not be strict.

7. thanks Isomorphism, i am familiar with the AM-GM inequality but i was hoping this could be pulled out by fiddling.

JaneBennet the Cauchy–Schwarz inequality is totally alien to me, you know any good article with an explanation / proof of it ?

8. Originally Posted by JaneBennet
They need to be positive for the inequality to be strict. For example, if $p=1,\ q=-1,\ r=1$, the inequality will not be strict.
q>r does not hold for your example.

That is p>q>r is enough, r>0 need not hold. Thats what I was trying to say

9. Originally Posted by Isomorphism
$a^2 + b^2 + c^2 \geq ab+bc+ca$
(By AM-GM, $\Sigma_{a,b,c}(a^2 + b^2) \geq \Sigma_{a,b,c}2ab$)
This is C-S not AM-GM.

10. Originally Posted by ThePerfectHacker
This is C-S not AM-GM.
No, I used AM-GM to get the result.Perhaps I should be more clear.

By AM-GM,

$(a^2 + b^2) \geq 2ab$
$(b^2 + c^2) \geq 2bc$
$(c^2 + a^2) \geq 2ca$

$2(a^2 + b^2 + c^2) \geq 2ab + 2bc + 2ca$

Cancel 2 on both sides to wrap it up
Yes I am aware that you can get it by a simple application of CS. But when we were in school we were taught only AM-GM. So I thought probably the poster wanted the proof using that

11. Or you can just use C-S:

$(a^2+b^2+c^2) \geq \sqrt{ab+bc+ac}\cdot \sqrt{ab+bc+ac}$