<s>[INST] Inequality $ab+bc+cd \geq 3abcd$ If $a,b,c,d$ are real numbers such that $a+b+c+d=6$ and $a^2+b^2+c^2+d^2=12$, then is this inequality true: $$ab+bc+cd\geq 3abcd$$
 [/INST] Let $$a=\frac{5}{2},b=\frac{1}{2}\left(\frac{3}{2}-\frac{\sqrt{5}}{2}\right),c=2,,d=\frac{1}{2}\left(\frac{3}{2}+\frac{\sqrt{5}}{2}\right)$$ we find $$a+b+c+d=6$$$$a^2+b^2+c^2+d^2=12$$ and $$ab+bc+cd=\frac{1}{8}\left(39-5\sqrt{5}\right)\approx 3.477$$ and $$3abcd=\frac{15}{4}=3.75$$ with $$ab+bc+cd<3abcd$$ so the inequality is not always true.
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