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.

Hint:
$$(a+b+c+d)^2=(a^2+b^2+c^2+d^2)+2(ab+ac+ad+bc+bd+cd)$$
Let
$$A=a+b+c+d$$
$$B=a^2+b^2+c^2+d^2$$
Now, this amounts to prove, 
$$\frac{A^2}{16} \geq \frac{A^2-B}{12}$$
$$4B \geq A^2$$
You can also write it
$$ \sqrt{a^2+b^2+c^2+d^2} \geq \frac{a+b+c+d}{2}$$
Or
$$ \sqrt{\frac{a^2+b^2+c^2+d^2}{4}} \geq \frac{a+b+c+d}{4}$$
And you should recognize this as a standard inequality between quadratic and arithmetic mean. You can prove it using Cauchy-Schwarz inequality, which states that
$$\left| \sum_{i=1}^n a_ib_i \right| \leq \sqrt{\sum_{i=1}^n a_i^2} \sqrt{\sum_{i=1}^n b_i^2}$$
With $b_i=1$ and $a_i \geq 0$, you get
$$ \sum_{i=1}^n a_i \leq \sqrt{\sum_{i=1}^n a_i^2} \sqrt{n}$$
Or
$$ \frac{1}{n}\sum_{i=1}^n a_i \leq \sqrt{\frac{1}{n}\sum_{i=1}^n a_i^2}$$

In case you don't know Cauchy-Schwarz inequality, here is a standard proof.
The following inequality is obviously true because it's a sum of squares:
$$\sum_{i=1}^{n} (a_i - \lambda b_i)^2 \geq 0$$
Now, develop w.r.t. $\lambda$, to get
$$ \left(\sum_{i=1}^n b_i^2\right) \lambda^2 - 2 \left(\sum_{i=1}^n a_ib_i\right) \lambda + \left(\sum_{i=1}^n a_i^2\right) \geq 0$$
It's a positive trinomial (in $\lambda$), thus its discriminant must be $\leq 0$ (otherwise there would be two roots, and it would becom negative between them, which is not possible).
Write the inequality on the discriminant:
$$4 \left(\sum_{i=1}^n a_ib_i\right)^2 - 4\left(\sum_{i=1}^n a_i^2\right)\left(\sum_{i=1}^n b_i^2\right) \leq 0$$
And finally,
$$ \left(\sum_{i=1}^n a_ib_i\right)^2 \leq \left(\sum_{i=1}^n a_i^2\right)\left(\sum_{i=1}^n b_i^2\right)$$
Or
$$\left| \sum_{i=1}^n a_ib_i \right| \leq \sqrt{\sum_{i=1}^n a_i^2} \sqrt{\sum_{i=1}^n b_i^2}$$
