# 2 nice inequalities

• Oct 3rd 2009, 05:29 AM
julia89
2 nice inequalities
hi. I'm not sure if this is a right topic but i have got problem with solving 2 inequalities:

(1) Numbers a,b,c are lengths of sides of a triangle, which square equals S, and let p,q,r be positive reals such that p+q+r=1. Prove that $\frac{p}{q+r}\cdot a^{2}+\frac{q}{p+r}\cdot b^{2}+\frac{r}{q+p}\cdot c^{2} \geqslant 2\sqrt{3}\cdot S$.

(2) Let a,b,c be positive reals, and n positive integer. Prove that $\frac{a^{n+1}}{b+c}+\frac{b^{n+1}}{a+c}+\frac{c^{n +1}}{b+a}\geqslant (\frac{a^{n}}{b+c}+\frac{a^{n}}{b+c}+\frac{a^{n}}{ b+c})\cdot (\frac{a^{n}+b^{n}+c^{n}}{3})^{\frac{1}{n}}$.

It is possible that it came from previous olympics (if its true its unlikely to be IMO, because I have already checked some of recent exercises. Then If you remember any of it just let me know source:)
• Oct 3rd 2009, 07:21 AM
red_dog
For 1) see this Topics in Inequalities Theorems and Techniques Hojolee
page 5, Theorem 6.
• Oct 3rd 2009, 01:47 PM
julia89
Thx a lot(Hi) I saw that Hadwiger-Finsler follows from Jensen applayed to function tangens.

Could you help me also with the second one?

• Oct 10th 2009, 02:25 PM
Marcel1289
I cant guarantee this, as I havent tried it personally, but for the first problem, you could try using the binomial expansion formula i.e. subtracting the right side and transforming it.

Im not sure if this will work though, so please do keep us informed on any progress (Happy)
• Oct 14th 2009, 11:59 AM
julia89
Could you explain your idea a little bit more?

I took both sides into n-th power and use following formula:
$(x+y+z)^{n}=\sum\limits_{i=0}^{n} ({n\choose i}a^{i}\cdot\sun\limits_{k=0}^{n-i} {n-i\choose k}b^{k}c^{n-i-k})$.

It look like a kind of rearragament inequality, but none of sequances I can choose are good ordered. I also tried to apply n-variables function version of it, but problem with the order doasnt disappear.

I know also general rearragament inequality, but dunno if could I apply it in the same way as in n-variables function version (still for lenght <=3)?
• Oct 14th 2009, 12:00 PM
julia89
Could you explain your idea a little bit more?

I took both sides into n-th power and use following formula:
$
(x+y+z)^{n}=\sum_{i=0}^{n} ({n\choose i}a^{i}\cdot\sum_{k=0}^{n-i} {n-i\choose k}b^{k}c^{n-i-k})
$
.

It look like a kind of rearragament inequality, but none of sequances I can choose are good ordered. I also tried to apply n-variables function version of it, but problem with the order doasnt disappear.

I know also general rearragament inequality, but dunno if could I apply it in the same way as in n-variables function version (still for lenght <=3)?