Question

Exercitiile 91, 96, 97. URGENT!!! DAU COROANA!!!

Answer 1

91.  1=a+b+c >=3 ·∛abc => 1 /abc >=27; 1 /√a +1 /√b +1 /√c >=3 ·1 /∛√a·√b·√c =3 /⁶√abc >=3√3 ;

96.  Se aplica inegalitatea mediilor (mg <=ma ) pentru n-1 numere reale pozitive si astfel obtinem

ⁿ⁻¹√a₂a₃...an +ⁿ⁻¹√a₁a₃...an +...+ⁿ⁻¹√a₁a₂...an₋₁ <=a₂+a₃+...+an /n-1 +a₁+a₃+...+an /n-1 +...+a₁+a₂+...+an₋1 =(n-1)(a₁+a₂+...+an) /n-1 dar deoarece ∑ a k=1 unde k=1,n (cu bara deasupra) => inegalitatea ceruta .

97.  1=ab+bc+ca >=3·∛ab·bc·ca => 1 /ab·bc·ca >=27 => 1 /aⁿ +1 /bⁿ +1 /cⁿ >=3· 1 /∛aⁿ·bⁿ·cⁿ =3 /⁶√ab·bc·ca =3√3ⁿ .