if $\displaystyle p > q > r > 0 $
prove $\displaystyle p^4 + q^4 + r^4 > p^2 q^2 +q^2 r^2 +r^2 p^2$
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.
$\displaystyle a^2 + b^2 + c^2 \geq ab+bc+ca$
(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)
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
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