Question

Daca a, b, c sunt numere reale pozitive, sa se demonstreze ca:
$a) 2( a^{3} + b^{3} + c^{3} ) \geq (a+b)ab+(b+c)bc+(a+c)ac$
$b) a^{2}+ b^{2} + c^{2} \geq ab+ac+bc$

Answer 1

[tex] $Aplicand inegalitatea mediilor numerelor $a^2,b^2$ obtinem:$\\ \frac{a^2+b^2}{2}\geq\sqrt{a^2b^2}=ab\Rightarrow a^2+b^2\geq2ab\Rightarrow \boxed{a^2+b^2-ab\geq ab} \\ a^3+b^3=(a+b)(a^2-ab+b^2)\geq(a+b)ab\\ a^3+c^3=(a+c)(a^2-ac+c^2)\geq(a+c)ac\\ c^3+b^3=(c+b)(c^2-cb+b^2)\geq(c+b)cb\\ $Sumand obtinem relatia dorita.$\\ [/tex]

[tex]a^2+b^2\geq2ab\\ c^2+b^2\geq2cb\\ a^2+c^2\geq2ac\\ $Sumam memebru cu membru:$\\ 2(a^2+b^2+c^2)\geq2(ab+bc+ac)\Leftrightarrow\\ a^2+b^2+c^2\geq ab+bc+ac\\ [/tex]