Hello I would like to solve this with $x,y,z$ positive real numbers :
$$\sqrt{\dfrac{5}{13}}\frac{1}{13}\dfrac{(x)^3}{(x^2+1)}+\dfrac{1}{13}\dfrac{(y)^2}{(y^2+1)}+\sqrt{\dfrac{13}{5}}\dfrac{1}{13(y)}\dfrac{(z)^2}{(z^2+1)}$$$$\geq \dfrac{1+\sqrt{\dfrac{5}{13}}x+\sqrt{\dfrac{13}{5}}\dfrac{1}{y}}{18}$$
under the condition $xyz=(\sqrt{\frac{13}{5}})^3$
I have no idea to prove this. Thanks a lot.
This answer is extremely technical and uses Mathcad calculations.
By routine calculations the inequality can be simplified to
$$\frac{5}{18}\left(\frac xA +1+\frac Ay\right)\ge \frac xA\cdot \frac 1{x^2+1}+\frac 1{y^2+1}+\frac Ay\cdot \frac 1{z^2+1},$$
where $A=\sqrt\frac{13}5$.
Or, because $z^2=\frac {A^6}{x^2y^2}$,
$$\frac{5}{18}\left(\frac xA +1+\frac Ay\right)\ge \frac xA\cdot \frac 1{x^2+1}+\frac 1{y^2+1}+\frac Ay\cdot \frac { x^2y^2}{A^6 + x^2y^2},$$
Substituting $u=\frac xA$, $v=\frac yA $ we obtain
$$\frac{5}{18}\left(u+1+\frac 1v\right)\ge u\cdot \frac 1{A^2u^2+1}+\frac 1{A^2v^2+1}+\frac 1v\cdot \frac {u^2v^2}{A^2 + u^2v^2}$$
$$u+1+\frac 1v\ge 18\left(u\cdot \frac 1{13u^2+5}+\frac 1{13v^2+5}+\frac 1v\cdot \frac {u^2v^2}{13 + 5u^2v^2}\right)$$
Substituting $w=\frac 1v$, we obtain
$$u+1+w\ge 18\left(\frac u{13u^2+5}+\frac {w^2}{13+5w^2}+ \frac {wu^2}{13w^2 + 5u^2}\right)$$
$25(u^5w^2+u^2+w^5)+65(u^5-u^4w^3-u^4w^2+u^4+u^3w^4-u^3+u^2w^5-u^2w-uw^4-w^4+w^3+w^2)+144 (u^3w^2+u^2w^3+u^2w^2)-169(u^4w+ u^2w^4+uw^2)\ge 0$
The graphs suggest that this inequality is true. We may look for its proof as follows.
Denote the left hand side of the last inequality by $f(u,w)$. We are going to show that $\inf\{f(u,w): u,w\ge 0\}=0$. When one of the variables $u$ and $w$ is fixed, $f(u,w)$ becomes a polynomial with respect to the other (which we denote by $v$) with the leading coefficient at least $25$. So it attains its minimum when $v=0$ or $\frac{\partial f}{\partial v}=0$. If $u=0$ then $f(u,w)=25w^5-65w^4+65w^3+65w^2=w^3(5w-8)^2+15w^4+w^3+65w^2\ge 0$. If $w=0$ then $f(u,w)=65u^5+65u^4-65u^3+25u^2=u^2(8u-5)^2+65u^5+u^4+15u^3\ge 0$.
Conditions $\frac{\partial f}{\partial u}=0$ and $\frac{\partial f}{\partial w}=0$ yield the system (*)
$130uw^5+(195u^2-338u-65)w^4+(-260u^3+288u)w^3+(125u^4-260u^3+432u^2+288u-169)w^2+(-676u^3-130u)w+ 325u^4+260u^3-195u^2+50u=0$
$(325u^2+125)w^4+(260u^3-676u^2-260u-260)w^3+(-195u^4+432u^2+195)+(50u^5-130u^4+288u^3+288u^2-338u+130)w+(-169u^4-65u^2)=0$
Its resultant (here) is a polynomial of $u$ of thirty second degree with many approximately twenty five digital integer coefficients. It has many real roots, among them $0$ and $1$ and the largest of them is approximately $2.3242305780688903970$. For the resultant which is a polynomial of $w$ we have a similar situation: matrix, polynomial, and roots. Assuming that the function $f$ attains its minimum at a point $(u,w)$ which is a solution of system (*), it remains to check the values $f(u,w)$ for each pair of these non-negative roots. It turned out that $f(0,0)=f(1,1)=0$ and $f(u,w)>0.02$ for any other pair of the roots. Thus assuming that my Mathcad calculated the roots with an error at most $4\cdot 10^{-9}$ , we have a proof.
COMMENT.-Noting $A=\sqrt{\dfrac{13}{5}}$ we have after some easy calculation the equivalent inequality $$xyf(x)+Ayf(y)+A^2f(z)\ge0$$ where $f(t)=\dfrac{t^2-A^2}{t^2+1}$ so $-A^2=-2,6\le f(t)\le0$ in the interval $[0,A]$ and $f(t)\gt0$ otherwise increasing till $1$ as limit when $t\to\infty$.
The fact that $f(A)=0$ makes easy the calculation with Lagrange multipliers: the minimum of the function $$F(x,y,z)= xyf(x)+Ayf(y)+A^2f(z)\\\text{ having }xyz=A^3\text{ as restriction }$$ This minimum is reached when $x=y=z=A$ and because of $F(A,A,A)=0$ we are done.
►However one wants to solve the problem with more basic mathematics. For example, by simple $AM\ge GM$ the form of $F(x,y,z)$ leads to the fact that the asked inequality is valid for all $(x,y,z)$ such that $\sqrt[3]{f(x)f(y)f(z)}\ge0$.
Because of $xyz=A^3$ the positive numbers $x,y,z$ can not be all the three less than $A$ nor greater than $A$ so at least one of them should be less than $A$. If one of the other two is less than $A$ we are done. Therefore it has been proved this way the inequality when two of the three variables are less than $A$.
► Consequently it remains to finish, the case in which two among $x,y,z$ are greater than $A$ (say $f(x)\gt0$ and $f(y)\gt 0$ and $f(z)\lt0$).
We left this part for the OP and just displayed the above solution with Lagrange multipliers.
