# Prove that the cyclic sum over $a, b, c > 0$ $\sum (\frac{a}{a+b})^3 >= \frac{3}{8}$

Prove that the cyclic sum over $$a, b, c > 0$$ $$\sum_{cyc} \left(\frac{a}{a+b}\right)^3 \geq \frac{3}{8}$$

I tried to do this in different way I am just given a sketch of what I did

first I tried to use holder Inequality $$\sum_{cyc} \left(\frac{a}{a+c}\right)^3 \cdot \sum_{cyc} \left(a\right)^3 \cdot \sum_{cyc} \left(\frac{1}{a+c}\right)^3 \geq 3$$

Then I tried to separate required equation out, but in that the inequality flipped.

I also tried to relate it to Nesbitt's inequality by first using $$\sum_{cyc} {\frac{a^3}{(a+c)^3}} \geq \frac{1}{9}\left(\sum_{cyc} {\frac{a}{a+c}}\right)^3$$ but now by rearrangement inequality the sign again get flipped. So both of these inequalities are quite weak,

I tried few more things like Cauchy–Schwarz inequality, AM-GM but all in vain for me, I think there would be a nice substitution but am not able to think of. Further it will be nice if someone give answer using holder though its not necessary (because I am practicing that)

Any hint or idea is Appreciated.

We can use the Vasc's inequality here: $$(a^2+b^2+c^2)^2\geq3(a^3b+b^3c+c^3a).$$ Indeed, by C-S we obtain: $$\sum_{cyc}\frac{a^3}{(a+b)^3}=\sum_{cyc}\frac{a^4}{a(a+b)^3}\geq$$ $$\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}a(a+b)^3}=\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^4+3a^3b+3a^2b^2+ab^3)}\geq$$ $$\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^4+3a^2b^2)+(a^2+b^2+c^2)^2+\frac{1}{3}(a^2+b^2+c^2)^2}\geq\frac{3}{8}.$$

• Nice, thanks a lot.
– pde
Jun 24, 2020 at 19:39
• Sorry for asking you so directly, put please have a look at this one. It seems that the question is related to inequalities so the answer might be already known to an expert on the subject like you. Jun 25, 2020 at 18:16
• @Maximilian Janisch Thank you for your respect to me. It seems very interesting. I'll see it later because I am very busy in last time. Jun 25, 2020 at 18:36
• @MichaelRozenberg No pressure 🙂 Jun 25, 2020 at 18:37
• @MichaelRozenberg FYI the problem is now solved :) Jun 26, 2020 at 15:56

Using the AM-GM inequality we have $$\frac{a^3}{(a+b)^3}+\frac{a^3}{(a+b)^3}+\frac{1}{8} \geqslant \frac{3}{2} \cdot \frac{a^2}{(a+b)^2}.$$ Therefore, we will show that $$\frac{a^2}{(a+b)^2}+\frac{b^2}{(b+c)^2}+\frac{c^2}{(c+a)^2} \geqslant \frac34.$$ According to the Cauchy-Schwarz inequality have $$\sum \frac{a^2}{(a+b)^2}=\sum \dfrac{a^2(a+c)^2}{(a+b)^2(a+c)^2} \geqslant \dfrac{\displaystyle \left[\sum (a^2+ab)\right]^2}{\displaystyle \sum (a+b)^2(a+c)^2}.$$ Finally, we need to prove $$4\left[\sum (a^2+ab)\right]^2 \geqslant 3 \sum (a+b)^2(a+c)^2,$$ equivalent to $$\left[(a+b)^2+(b+c)^2+(c+a)^2\right]^2 \geqslant 3\sum (a+b)^2(a+c)^2.$$ This is $$(x+y+z)^2 \geqslant 3(xy+yz+zx).$$

• Nice, thanks a lot
– pde
Jun 25, 2020 at 14:06