<s>[INST] showing that $\frac{a+b}{2}\cdot\frac{c+d}{2} =\big( \frac{a+b+c+d}{4}\big)^2$ I could expand the right side of the equation below to show that the sides are equal but I was wondering if there was some equality that would have allowed me to see that the left side equals the right?  and $a,b,c,d >0$
$$ = \frac{a+b}2 \cdot \frac{c+d}2 \cdot \frac{a+b+c+d}4 = \Big( \frac{a+b+c+d}4 \Big)^3 $$
Here's the entire proof.  The problem is from Arthur Engel's problem solving strategies in the section on inequalities.  The question was to prove the inequality below
$3$. We have
$$ \frac{abc + abd + acd + bcd}4 = \frac12 \Big( ab \frac{c+d}2 + cd \frac{a+b}2 \Big) $$
$$ \le \frac12 \Big( \Big( \frac{a+b}2 \Big)^2 \frac{c+d}2 + \Big( \frac{c+d}2 \Big)^2 \frac{a+b}2 \Big) $$
$$ = \frac{a+b}2 \cdot \frac{c+d}2 \cdot \frac{a+b+c+d}4 \le \Big( \frac{a+b+c+d}4 \Big)^3. $$
Hence,
$$ \sqrt[3]{ \frac{abc + abd + acd + bcd}4 } \le \frac{a+b+c+d}4 \le \sqrt{ \frac{a^2+b^2+c^2+d^2}4}. $$
 [/INST] Let


*

*$x=\frac{a+b}{2}$

*$y= \frac{c+d}{2}$


then
$$\frac{a+b}{2}\cdot\frac{c+d}{2} =\big( \frac{a+b+c+d}{4}\big)^2\iff xy=\left(\frac{x+y}{2}\right)^2$$
which is false in general indeed by AM-GM
$$\frac{x+y}{2}\ge \sqrt{xy} \iff xy\le \left(\frac{x+y}{2}\right)^2$$
and equality holds if and only if $x=y$.
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