Inegalitate conditionata...(Conc,Ungaria-2014)

[mihai miculita]

Aratati ca: $abc=1\Rightarrow\dfrac{a^3+b^3}{a^2+ab+b^2}+\dfrac{a^3+c^3}{a^2+ac+c^2}$$+\dfrac{b^3+c^3}{b^2+bc+c^2}\ge 2;\,(\forall)a,b,c>0.$

[DanDumitrescu]

Vom folosi urmatorul truc: $\dfrac{a^3+b^3}{a^2+ab+b^2}=$$a+b-\dfrac{2ab(a+b)}{a^2+ab+b^2}.$ Acum membrul intai al relatiei din enunt va fi egal cu:
$2(a+b+c)-\dfrac{2ab(a+b)}{a^2+ab+b^2}-\dfrac{2bc(b+c)}{b^2+bc+c^2}-\dfrac{2ca(c+a)}{c^2+ca+a^2}\ge$$\ 2(a+b+c)-\dfrac{2ab(a+b)}{3ab}-$$\dfrac{2bc(b+c)}{3bc}-$$\dfrac{2ca(c+a)}{3ca}=$$2.\dfrac{a+b+c}{3}\ge$$2.\sqrt[3]{abc}=2.$

[dangerous storm]

Alta solutie:
Se observa cu usurinta ca $\dfrac{a^3+b^3}{a^2+ab+b^2}\ge\dfrac{a+b}{3}$,deci $\sum_{cyc}\dfrac{a^3+b^3}{a^2+ab+b^2}\ge$$\dfrac{a+b}{3}+\dfrac{b+c}{3}+\dfrac{c+a}{3}=2\cdot\dfrac{a+b+c}{3}\ge 2\cdot \sqrt[3]{abc}=2$.