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 $a+b+c+d = 1$?

Question

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!

Answer 1

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, $$\sum\limits_{cyc} \frac{a^2}{a+b} = \frac{1}{2}\sum\limits_{cyc} \frac{a^2+b^2}{a+b} \ge \frac{1}{4}\sum\limits_{cyc} \frac{(a+b)^2}{a+b} = \frac{1}{2}\sum\limits_{cyc} a$$

Answer 2

Another approach using $4ab\le(a+b)^2$: \begin{align*}\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a}&=\left(a-\frac{ab}{a+b}\right)+\left(b-\frac{bc}{b+c}\right)+\left(c-\frac{cd}{c+d}\right)+\left(d-\frac{da}{d+a}\right)\\ &=1-\left(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{cd}{c+d}+\frac{da}{d+a}\right)\\ &\ge1-\frac14\Bigl((a+b)+(b+c)+(c+d)+(d+a)\Bigr)=\frac12\end{align*}

Answer 3

Use Cauchy-Schwarz: $$ \frac{a^2}{a+b}+\frac{d^2}{a+d}+\frac{b^2}{b+c}+\frac{c^2}{c+d}=\left(\frac{a^2}{a+b}+\frac{d^2}{a+d}+\frac{b^2}{b+c}+\frac{c^2}{c+d}\right)\cdot\left((a+b)+(a+d)+(b+c)+(c+d)\right)\cdot 0.5\geq (a+b+c+d)^2\cdot 0.5= 0.5 $$