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Prove that for any $a, b \in \mathbb{R}$ the following inequality holds true: $$a^2+b^2+16\geq ab+4a+4b$$ I tried some reductions, but with no success. Any hint appreciable! Thanks in advance.

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  • $\begingroup$ $a^2+b^2+c^2\ge ab+bc+ac$ $\endgroup$ – chenbai Sep 23 '14 at 8:50
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$$4a^2+4b^2+64-4ab-16a-16b=(a+b-8)^2+3(a-b)^2$$

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