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