Question

Am nevoie de ajutor la exercițiul 9.Va rog.e urgent

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

$\it b)\\ \\ (a-b)^2\geq0\ \ \ \ (1)\\ \\ (b-c)^2\geq0\ \ \ \ (2)\\ \\ (c-a)^2\geq0\ \ \ \ (3)\\ \\ (1),\ (2),\ (3) \Rightarrow (a-b)^2+(b-c)^2+(c-a)^2\geq0 \Rightarrow \\ \\  a^2-2ab+b^2+b^2-2ac+c^2+c^2-2ac+a^2\geq0 \Rightarrow \\ \\ \Rightarrow 2a^2+2b^2+2c^2\geq 2ab+2bc+2ca \Rightarrow2(a^2+b^2+c^2)\geq2(ab+bc+ca)|_{:2}\Rightarrow\\ \\ \Rightarrow a^2+b^2+c^2\geq ab+bc+ca$

( Egalitatea are loc dacă a = b = c ) !

$\it c)\ \ a^2+b^2+c^2=ab+bc+ca \Rightarrow a=b=c\ \ \ \ (*)\\ \\ a+2b+3c=18\sqrt2 \stackrel{(*)}{\Longrightarrow}\ a+2a+3a=18\sqrt2 \Rightarrow 6a=18\sqrt2|_{:6}\Rightarrow a=3\sqrt2\\ \\ Deci,\ \ a=b=c=3\sqrt2$