Ytukyg

Let $a,b,c$ be positive real numbers. Prove that $$Big({aover a+b}Big)^3+Big({bover b+c}Big)^3+ Big({cover c+a}Big)^3geq {3over 8}$$

If we put $x=b/a$, $y= c/b$ and $z=a/c$ we get $xyz=1$ and

$$Big({1over 1+x}Big)^3+Big({1over 1+y}Big)^3+ Big({1over 1+z}Big)^3geq {3over 8}$$

Since $f(x)=Big({1over 1+x}Big)^3$ is convex we get, by Jensen,: $$Big({1over 1+x}Big)^3+Big({1over 1+y}Big)^3+ Big({1over 1+z}Big)^3geq 3f({x+y+zover 3})$$

Unfortunately, since $f$ is decreasing we don't have $f({x+y+zover 3}) geq f(1) = {1over 8}$.

Some idea how to solve this?

Let $g(t):=left(dfrac{1}{1+exp(t)}right)^3$ for $tinmathbb{R}$. Then, $$g''(t)=frac{3,exp(t),big(3,exp(t)-1big)}{big(1+exp(t)big)^5}text{ for each }tinmathbb{R},.$$
Thus, $g$ is convex on $big[-ln(3),inftybig)$.

Let $x:=dfrac{b}{a}$, $y:=dfrac{c}{b}$, and $z:=dfrac{a}{c}$ be as the OP defines. Then, the required inequality is equivalent to
$$gbig(ln(x)big)+gbig(ln(y)big)+gbig(ln(z)big)geq dfrac{3}{8},.tag{*}$$
If $x$, $y$, or $z$ is less than $dfrac{1}{3}$, then clearly the left-hand side of (*) is greater than
$$left(dfrac{1}{1+frac13}right)^3=frac{27}{64}>frac38,.$$
If all $x$, $y$, and $z$ are greater than or equal to $dfrac13$, then $ln(x),ln(y),ln(z)geq -ln(3)$, so that we can use convexity of $g$ on $big[-ln(3),inftybig)$. By Jensen's Inequality,
$$gbig(ln(x)big)+gbig(ln(y)big)+gbig(ln(z)big)geq3,gleft(frac{ln(x)+ln(y)+ln(z)}{3}right)=3,g(0)=frac{3}{8},.$$
Hence, the equality holds if and only if $x=y=z=1$, making $a=b=c$.

By Holder $$left(sum_{cyc}frac{a^3}{(a+b)^3}right)^2sum_{cyc}1geqleft(sum_{cyc}sqrt[3]{left(frac{a^3}{(a+b)^3}right)^2cdot1}right)^3=left(sum_{cyc}frac{a^2}{(a+b)^2}right)^3.$$
Thus, it's enough to prove that
$$frac{left(sumlimits_{cyc}frac{a^2}{(a+b)^2}right)^3}{3}geqfrac{9}{64}$$ or
$$sumlimits_{cyc}frac{a^2}{(a+b)^2}geqfrac{3}{4}.$$
Now, by C-S
$$sumlimits_{cyc}frac{a^2}{(a+b)^2}=sumlimits_{cyc}frac{a^2(a+c)^2}{(a+b)^2(a+c)^2}geqfrac{left(sumlimits_{cyc}(a^2+ab)right)^2}{sumlimits_{cyc}(a+b)^2(a+c)^2}.$$
Thus, it's enough to prove that
$$4left(sumlimits_{cyc}(a^2+ab)right)^2geq3sumlimits_{cyc}(a+b)^2(a+c)^2,$$
which is true even for all reals $a$, $b$ and $c$.

Indeed, the last inequality is symmetric inequality by degree four,

which says that by $uvw$ (https://math.stackexchange.com/tags/uvw/info )

it's enough to prove the last inequality for equality case of two variables and since

it's the homogeneous inequality by even degree, we can assume $b=c=1$, which gives
$$(a-1)^2(a+3)^2geq0.$$
Done!

This is more of a comment, but I don't have the reputation. Use Lagrange multipliers. Solving, you find that the critical points occur when $xyz=1$ and $yz (1+x)^4 = xz (1+y)^4 = xy (1+z)^4$. I think the only solution is $x=y=z=1$. Clearly, it's a minimum and plugging in shows the bound.

Added:

We can rewrite the condition as $xyz=1$ and $frac{(1+x)^4}{x} = frac{(1+y)^4}{y} = frac{(1+z)^4}{z}$

The function $g(x)=frac{(1+x)^4}{x}$ is decreasing from $0$ to $1/3$ and increasing from $1/3$ to $infty$. This shows that 2 of $x,y,z$ must be equal (WLOG $x$ and $y$) and $z=1/x^2$. It remains to solve $frac{(1+x)^4}{x} = frac{(1+1/x^2)^4}{1/x^2}$. This time, it's not hard to check $x=1$ is the only solution and we are done.