parmenides51 wrote: For arbitrary positive integers $a\ge b \ge c$ prove the inequality: $$\frac{a^2+b^2}{a+b}+\frac{a^2+c^2}{a+c}+\frac{c^2+b^2}{c+b}\ge (a+b+c)+ \frac{(a-c)^2}{a+b+c}$$(Anton Trigub) It's true for any positives $a$, $b$ and $c$. C-S of course.

bad translation by google, reals and not integers is the correct

Solution:

Let $a,b,c$ be positive reals. Prove that $$\frac{a^2+b^2}{a+b}+\frac{a^2+c^2}{a+c}+\frac{c^2+b^2}{c+b}\ge a+b+c + \frac{(a-c)^2}{a+b+c}$$Solution: $$\sum_{cyc}\frac{a^2+b^2}{a+b}= \sum_{cyc}a+\frac{1}{2}\sum_{cyc}\frac{(a-b)^2}{a+b}\geq \sum_{cyc}a+\frac{1}{2}\left( \frac{(a-c)^2}{a+2b+c}+\frac{(c-a)^2}{c+a}\right)\ge \sum_{cyc}a+\frac{(a-c)^2}{a+b+c}$$

parmenides51 wrote: For arbitrary positive reals $a\ge b \ge c$ prove the inequality: $$\frac{a^2+b^2}{a+b}+\frac{a^2+c^2}{a+c}+\frac{c^2+b^2}{c+b}\ge (a+b+c)+ \frac{(a-c)^2}{a+b+c}$$(Anton Trygub) $$\sum \frac{a^2+b^2}{a+b} - \sum \frac{a+b}{2} = \frac{(a-b)^2}{2(a+b)} + \frac{(b-c)^2}{2(b+c)} + \frac{(a-c)^2}{2(c+a)} \ge \frac{(2a-2c)^2}{4(a+b+c)} = \frac{(a-c)^2}{a+b+c}$$$\blacksquare$

For arbitrary positive reals $a, b,c$ prove the inequality: $$\frac{a^3+b^3}{a+b}+\frac{a^3+c^3}{a+c}+\frac{c^3+b^3}{c+b}\ge a^2+b^2+c^2+ \frac{6(a-c)^4}{7(a^2+b^2+c^2)}$$$$\sqrt{\frac{a^3+b^3}{a+b}}+\sqrt{\frac{a^3+c^3}{a+c}}+\sqrt{\frac{c^3+b^3}{c+b}}\ge a+b+c+ \frac{(a-c)^2}{a+b+c}$$https://artofproblemsolving.com/community/c6h1791421p11852916