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Conditionata cu raport de suma de cuburi
Posted: Wed Jun 10, 2009 3:04 pm
by alex2008
Fie \( a,b,c,d\ge 0 \) astfel incat \( \frac{a^3+b^3}{a+b}=\frac{c^3+d^3}{c+d} \) . Sa se demonstreze ca :
\( (a+b)(c+d)\ge 2(ab+cd) \)
Posted: Wed Jun 10, 2009 4:55 pm
by Mateescu Constantin
Conditia din enunt este echivalenta cu \( a^2-ab+b^2=c^2-cd+d^2 \)
Fie \( x=a^2-ab+b^2=c^2-cd+d^2 \). Fara a restrande generalitatea sa presupunem ca \( ab\ge cd \).
\( \Longrightarrow \left\{ \begin x\ge ab\ge cd\\ (a+b)^2=x+3ab \\ (c+d)^2=x+3cd \)
Ridicand la patrat inegalitatea de demonstrat devine
\( (x+3ab)(x+3cd)\ge 4(ab+cd)^2 \)
\( \Longrightarrow (x+3ab)(x+3cd)-4(ab+cd)^2\ge 4ab(ab+3cd)-4(ab+cd)^2=4cd(ab-cd)\ge 0 \)
Egalitatea are loc pentru \( (a,\ b,\ c,\ d)\sim (1,\ 1,\ 1,\ 1) \) si \( (a,\ b,\ c,\ d)\sim (0,\ 1,\ 1,\ 1) \) sau permutarile lor ciclice.