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

Question

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.

Thanks in advance.

Answer 1

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: $$\sum_{cyc}\frac{a^3}{(a+b)^3}=\sum_{cyc}\frac{a^4}{a(a+b)^3}\geq$$ $$\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}a(a+b)^3}=\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^4+3a^3b+3a^2b^2+ab^3)}\geq$$ $$\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^4+3a^2b^2)+(a^2+b^2+c^2)^2+\frac{1}{3}(a^2+b^2+c^2)^2}\geq\frac{3}{8}.$$

Answer 2

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 $(x+y+z)^2 \geqslant 3(xy+yz+zx).$