Question

Vă rog ajutați-mă cu ultimul .

Answer 1

[tex]\it Fie \ a,\ b,\ c \in \mathbb{R}.\ Aplic\ inegalitatea\ mediilor,\ astfel: \\\;\\ \dfrac{a^2+b^2}{2} \geq\sqrt{a^2\cdot b^2} \Rightarrow \dfrac{a^2+b^2}{2} \geq |a|\cdot|b| \Rightarrow \dfrac{a^2+b^2}{2} \geq ab \ \ \ (1) [/tex]


[tex]\it \dfrac{b^2+c^2}{2} \geq\sqrt{b^2\cdot c^2} \Rightarrow \dfrac{b^2+c^2}{2} \geq |b|\cdot|c| \Rightarrow \dfrac{b^2+c^2}{2} \geq bc \ \ \ (2) \\\;\\ \\\;\\ \dfrac{c^2+a^2}{2} \geq\sqrt{c^2\cdot a^2} \Rightarrow \dfrac{c^2+a^2}{2} \geq |c|\cdot|a| \Rightarrow \dfrac{c^2+a^2}{2} \geq ca \ \ \ (3)[/tex]

Prin însumarea relațiilor (1), (2), (3), rezultă:

[tex]\it \dfrac{a^2+b^2+b^2+c^2+c^2+a^2}{2} \geq ab+bc+ca \Rightarrow \\\;\\ \\\;\\ \Rightarrow \dfrac{2(a^2+b^2+c^2)}{2} \geq ab+bc+ca \Rightarrow a^2+b^2+c^2\geq ab+bc+ca [/tex]