[tex]\it Fie \ a,\ b,\ c \in \mathbb{R}.\ Aplic\ inegalitatea\ mediilor,\ astfel:
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\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)
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[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)
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\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
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\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]
