Răspuns:
[tex]a) \\a^2+b^2\geq 2ab\\(a+b)^2-2ab\geq 2ab\\(a+b)^2\geq 0[/tex]"Adevarat"
Explicatie:
[tex](a+b)^2=a^2+b^2+2ab=>a^2+b^2=(a+b)^2-2ab[/tex]
[tex]b) a^2+b^2+c^2\geq ab+bc+ca\\a^2+b^2+c^2-ab-bc-ca\geq 0|*(2)\\2a^2+2b^2+2c^2-2ab-2b-2ca\geq 0\\(a-b)^2+(a-c)^2+(b-c)^2\geq 0[/tex]"Adevarat"
[tex]c) a+2b+3c=18\sqrt{2}[/tex]
[tex]a^2+b^2+c^2=ab+bc+ca|*(2)\\2a^2+2b^2+2c^2-2ab-2bc-2ca=0\\a^2-2ab+b^2+b^2-2bc+c^2+a^2-2ca+c^2=0\\(a-b)^2+(b-c)^2+(c-a)^2=0\\a=b=c\\a+2b+3c=18\sqrt{2}\\b+2b+3b=18\sqrt{2}\\6b=18\sqrt{2}=>b=3\sqrt{2}=a=c[/tex]
