<s>[INST] Algebra Inequality Proof Let $a, b, c, d$ be positive numbers. Prove that $\frac{a+b+c+d}{4} \ge \sqrt{\frac{ab+ac+ad+bc+bd+cd}{6}}$.
I was told to rewrite the sum on the right side in terms of $a^2 + b^2 + c^2 + d^2$ and $a + b + c + d$ but I am unsure how to combine the terms.
 [/INST] another approach is: square both sides and rearrange, we have:
$\iff3(a^2+b^2+c^2+d^2)\ge 2(ab+ac+ad+bc+bd+cd)$
with $a^2+b^2+c^2\ge ab+bc+ac$ <1>
we also have:   $a^2+b^2+d^2\ge ab+bd+ad$<2> and $a^2+c^2+d^2\ge ad+cd+ac$<3>
edit: $b^2+c^2+d^2\ge bc+bd+cd$ <4>
<1>+<2>+<3>+<4> $\implies 3(a^2+b^2+c^2+d^2)\ge 2(ab+ac+ad+bc+bd+cd)$
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