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vansang02121998

$a^2+\dfrac{3-\sqrt{5}}{2}b^2 \ge ab(\sqrt{5}-1)$

$a^2+\dfrac{3-\sqrt{5}}{2}c^2 \ge ac(\sqrt{5}-1)$

$\dfrac{\sqrt{5}-1}{2}b^2+\dfrac{\sqrt{5}-1}{2}c^2 \ge bc(\sqrt{5}-1)$

$\Rightarrow 2a^2+b^2+c^2 \ge (ab+ac+bc)(\sqrt{5}-1)$
 
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