Inequality rebirth!! ( + difficile ...)

[Simo_the_wolf]

$ a+b+c=0 $ con $ abc \neq 0 $.

Dimostrare che:

1) $ \displaystyle \frac {a^2}{b^2} + \frac {b^2}{c^2} + \frac {c^2}{a^2} \geq 5 $

2) $ \displaystyle \frac {a^2}{b^2} + \frac {b^2}{c^2} + \frac {c^2}{a^2} + $ $ \displaystyle \frac {b^2}{a^2} + \frac {c^2}{b^2} + \frac {a^2}{c^2} \geq 10 + \frac 12 $

3) $ \displaystyle \frac {a^4}{b^4} + \frac {b^4}{c^4} + \frac {c^4}{a^4} > 12 + \frac 12 $

[darkcrystal]

La lezione del coraggio
1. Facendo il minimo comune multiplo e i "conticini", si trova, dopo aver sostituito c=-a-b, $ \displaystyle \frac{(a^3+a^2b-2ab^2-b^3)^2}{a^2b^2(a+b)^2} \geq 0 $. Se non vi fidate contate

Ciao!

[post233]

2-) $ \displaystyle \frac{a^2}{b^2}+\displaystyle \frac{b^2}{c^2}+\displaystyle \frac{c^2}{a^2}+\displaystyle \frac{a^2}{c^2}+\displaystyle \frac{b^2}{a^2}+\displaystyle \frac{c^2}{b^2} \geq \displaystyle \frac{21}{2} $
Sappiamo che $ c=-a-b $, dunque $ LHS=\displaystyle \frac{a^2}{b^2}+\displaystyle \frac{b^2}{(a+b)^2}+\displaystyle \frac{(a+b)^2}{a^2}+\displaystyle \frac{a^2}{(a+b)^2}+\displaystyle \frac{b^2}{a^2}+\displaystyle \frac{(a+b)^2}{b^2}= $
$ =2\displaystyle \frac{a^2}{b^2}+2\displaystyle \frac{b^2}{a^2}+1-\displaystyle \frac{2ab}{(a+b)^2}+1+\displaystyle \frac{2ab}{a^2}+1+\displaystyle \frac{2ab}{b^2}= $
$ 2(\displaystyle \frac{a^2}{b^2}+\displaystyle \frac{b^2}{a^2}+2)-1+2\displaystyle \frac{a}{b}+2 \displaystyle \frac{b}{a}-\displaystyle \frac{2ab}{(a+b)^2}= $
$ =2(\displaystyle \frac{b}{a}+\displaystyle \frac{a}{b})^2+2(\displaystyle \frac{b}{a}+\displaystyle \frac{a}{b})-\displaystyle \frac{[G.M.(a,b)]^2}{2[A.M.(a,b)]^2}-1 \geq 8+4-\displaystyle \frac{1}{2}-1=\displaystyle \frac{21}{2} $
($ \displaystyle \frac{a}{b}+\displaystyle \frac{b}{a} \geq 2 $ per riarrangiamento).