<s>[INST] Why is $ \frac{a^2}{a+b}+\frac{d^2}{a+d}+\frac{b^2}{b+c}+\frac{c^2}{c+d} \geq 0.5 $ with $a+b+c+d = 1$? For positive real numbers $a,b,c,d>0$ it seems to be true that:
if $$a+b+c+d = 1$$
then 
$$  \frac{a^2}{a+b}+\frac{d^2}{a+d}+\frac{b^2}{b+c}+\frac{c^2}{c+d} \geq 0.5  $$
I can't think of a way to prove this statement.
Any help will be appreciated!
 [/INST] Use Cauchy-Schwarz Inequality: $$(a+b+b+c+c+d+d+a)\left(\frac{a^2}{a+b}+\frac{d^2}{a+d}+\frac{b^2}{b+c}+\frac{c^2}{c+d}\right) \ge (a+b+c+d)^2$$
Alternative approach: $$\sum\limits_{cyc} \frac{a^2}{a+b} = \sum\limits_{cyc} \left(\frac{a^2}{a+b} - (a-b)\right) = \sum\limits_{cyc} \frac{b^2}{a+b}$$
Therefore, $$\sum\limits_{cyc} \frac{a^2}{a+b} = \frac{1}{2}\sum\limits_{cyc} \frac{a^2+b^2}{a+b} \ge \frac{1}{4}\sum\limits_{cyc} \frac{(a+b)^2}{a+b} = \frac{1}{2}\sum\limits_{cyc} a$$
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