$\sum \frac{a}{b+c+d}\le \frac{2\sum a^2}{\sum ab}$ if $\sum a =4$ Let $a, b, c, d$ positive real numbers such that $a+b+c+d=4$. Prove that
$$\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{d+a+b}+\frac{d}{a+b+c}\le
\frac{2(a^2+b^2+c^2+d^2)}{ab+ac+ad+bc+bd+cd}.$$
My idea is to cancel the denominators, expand the inequality and use Muirhead, but the calculations are terrible... So I'm looking for a smarter solution.

Fill in any missing gaps.

*

*Multiplying by $(ab+ac+ad+bc+bd+cd)$, show that the inequality is equivalent to  $$\sum \frac{ abc+abd+acd}{b+c+d} \leq \sum a^2. $$

*Show that $$ \frac{ abc+abd+acd}{b+c+d} \leq \frac{ab+ac+ad}{3} \leq \frac{3a^2+b^2+c^2+d^2}{6}$$

*Hence, summing up the cyclic inequalities, the conclusion follows.


Note: This could be combined into 1 (non-obvious) step by showing that
$$ \frac{a}{b+c+d} \leq \frac{ 3a^2 + b^2+c^2+d^2} { 3(ab+ac+ad+bc+bd+cd)}.$$
