if $\displaystyle p > q > r > 0 $

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

Printable View

- Dec 29th 2007, 05:14 AMbobakInequality
if $\displaystyle p > q > r > 0 $

prove $\displaystyle p^4 + q^4 + r^4 > p^2 q^2 +q^2 r^2 +r^2 p^2$ - Dec 29th 2007, 05:29 AMPlato
$\displaystyle p > q > 0\; \Rightarrow \;p^2 > pq\; \Rightarrow \;p^4 > p^2 q^2 $

Can you use this to complete the problem? - Dec 29th 2007, 06:24 AMbobak
- Dec 29th 2007, 06:50 AMJaneBennet
Use the Cauchy–Schwarz inequality:

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

Equality obtains if and only if the vectors $\displaystyle \left(\begin{array}{c}p^2\\q^2\\r^2\end{array}\rig ht)$ and $\displaystyle \left(\begin{array}{c}q^2\\r^2\\p^2\end{array}\rig ht)$ in $\displaystyle \mathbb{R}^3$ are linearly dependent; i.e. if and only if $\displaystyle p^2=q^2=r^2$. Since $\displaystyle p>q>r>0$, this won’t happen and so the inequality is always strict. - Dec 29th 2007, 07:00 AMIsomorphism
$\displaystyle a^2 + b^2 + c^2 \geq ab+bc+ca$

(By AM-GM, $\displaystyle \Sigma_{a,b,c}(a^2 + b^2) \geq \Sigma_{a,b,c}2ab$)

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

The above AM-GM holds iff a=b=c.

Or if $\displaystyle p^2=q^2=r^2$

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):p - Dec 29th 2007, 07:13 AMJaneBennet
- Dec 29th 2007, 07:33 AMbobak
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 ? - Dec 29th 2007, 07:34 AMIsomorphism
- Dec 29th 2007, 05:34 PMThePerfectHacker
- Dec 29th 2007, 06:10 PMIsomorphism
No, I used AM-GM to get the result.Perhaps I should be more clear.

By AM-GM,

$\displaystyle (a^2 + b^2) \geq 2ab$

$\displaystyle (b^2 + c^2) \geq 2bc$

$\displaystyle (c^2 + a^2) \geq 2ca$

Add them up,

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

Cancel 2 on both sides to wrap it up (Whew)

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 :) - Dec 29th 2007, 06:14 PMThePerfectHacker
Or you can just use C-S:

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