<s>[INST] two inequalities of four real numbers If $a,b,c,d$ are four positive real numbers, then show that
   $$\frac{12}{(a+b+c+d)} \leq { \frac{1}{a+b} + \frac{1}{a+c} + \frac{1}{a+d} + \frac{1}{b+c} + \frac{1}{b+d} + \frac{1}{c+d}} \leq \frac34 \left[\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}\right]$$
 [/INST] Hint: $\frac{4}{x+y}\leq \frac{1}{x}+\frac{1}{y}$ for $x\gt0$ and $y\gt0$ 
To prove the hint, notice it's equivalent to $\frac{4}{x+y}\leq \frac{x+y}{xy}$, cross multiply(you can do it because both $x$ and $y$ are positive) to get the equivalent inequality $(x+y)^2\geq4xy$, which is proved in the comment.
Back to original question, for the first one, use the hint
$\frac{4}{a+b+c+d}\leq\frac{1}{a+b}+\frac{1}{c+d}$, write down the other two similiar inequalities, add them together. For the second one $\frac{4}{a+b}\leq\frac{1}{a}+\frac{1}{b}$, write down similar inequalities and add them together.
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