Lukaluce wrote: Let $x, y,$ and $z$ be positive real numbers such that $xy + yz + zx = 3$. Prove that $$\frac{x + 3}{y + z} + \frac{y + 3}{z + x} + \frac{z + 3}{x + y} \ge 27 \cdot \frac{(\sqrt{x} + \sqrt{y} + \sqrt{z})^2}{(x + y + z)^3}.$$ Proposed by Petar Filipovski, Macedonia No.

Let $x, y,$ and $z$ be positive real numbers such that $xy + yz + zx = 3$. Prove that $$\frac{x + 3}{y + z} + \frac{y + 3}{z + x} + \frac{z + 3}{x + y} \ge18 \cdot \frac{(\sqrt{x} + \sqrt{y} + \sqrt{z})^2}{(x + y + z)^3} $$Maybe.

Lukaluce wrote: Let $x, y,$ and $z$ be positive real numbers such that $xy + yz + zx = 3$. Prove that $$\frac{x + 3}{y + z} + \frac{y + 3}{z + x} + \frac{z + 3}{x + y} \ge 27 \cdot \frac{(\sqrt{x} + \sqrt{y} + \sqrt{z})^2}{(x + y + z)^3}.$$ Proposed by Petar Filipovski, Macedonia Yes.

Lukaluce wrote: Let $x, y,$ and $z$ be positive real numbers such that $xy + yz + zx = 3$. Prove that $$\frac{x + 3}{y + z} + \frac{y + 3}{z + x} + \frac{z + 3}{x + y} + 3 \ge 27 \cdot \frac{(\sqrt{x} + \sqrt{y} + \sqrt{z})^2}{(x + y + z)^3}.$$

Attachments:

Let $x+y+z=s$. By Cauchy-Schwarz, $\frac{x + 3}{y + z} + \frac{y + 3}{z + x} + \frac{z + 3}{x + y} \geq \frac{(x+y+z+9)^2}{\displaystyle \sum(x+3)(y+z)}=\dfrac{(s+9)^2}{6s+6},$ and by Cauchy-Schwarz again $27 \cdot \frac{(\sqrt{x} + \sqrt{y} + \sqrt{z})^2}{(x + y + z)^3} \leq \dfrac{27 \cdot 3s}{s^3}=\dfrac{81}{s^2},$ hence we are left to prove that $\dfrac{(s+9)^2}{6s+6}+3 \geq \dfrac{81}{s^2},$ which rewrites as $(s-3)(s^3+39s^2+216s+162) \geq 0,$ which is true as $s^2 \geq 3(xy+yz+zx)=9,$ that is $s \geq 3$.

this problem was in the 1st selection exam of Azerbaijan

Even easier way is to see $\sum \frac{x}{y+z}\ge \frac32$ (Nesbitt's) and $\sum \frac{3}{x+y}\ge \frac{27}{2(x+y+z)}$ (AM-HM), so setting $s=x+y+z$ as Orestis did above it boils down proving $\frac92+ \frac{27}{2s}\ge \frac{81}{s^2}$ (where I used $3s\ge (\sqrt{x}+\sqrt{y}+\sqrt{z})^2$ by Cauchy-Schwarz). But this is equivalent to $(s-3)(s+6)\ge 0$ and clearly $s^2\ge 3(xy+yz+zx)=9$ so $s\ge 3$, which yields the conclusion.

Rewrite the given inequality as \[(x+y+z+3)\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\right) \geq\dfrac{27(\sqrt{x}+\sqrt{y}+\sqrt{z})^2}{(x+y+z)^3}.\]By C-S inequality, we have \[\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}=\dfrac{z}{zx+zy}+\dfrac{x}{xy+xz}+\dfrac{y}{yz+yx}\geq \dfrac{(\sqrt{x}+\sqrt{y}+\sqrt{z})^2}{2(xy+yz+zx)}=\dfrac{(\sqrt{x}+\sqrt{y}+\sqrt{z})^2}{6},\]so it suffices to prove that $\dfrac{x+y+z+3}{6}\geq \dfrac{27}{(x+y+z)^3}$. Setting $t := x+y+z\geq 3$ which follows from AM-GM inequality, the last inequality is equivalent to $t^3(t+6)\geq 162\iff (t-3)(t^3+6t^2+18t+54)\geq 0$, which is clearly true. Equality holds when $x=y=z=1$.

We have, $$(x+y+z)^2\ge 3(xy+yz+zx)=9 \Rightarrow x+y+z\ge3$$Therefore, $$(x+y+z)^2\ge 3(x+y+z)\ge (\sqrt{x}+\sqrt{y}+\sqrt{z})^2 \Rightarrow (x+y+z)^2\ge (\sqrt{x}+\sqrt{y}+\sqrt{z})^2$$and $$x +y +z +3\ge6$$By Titu's, we have $$(x + y+ z)\left(\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}\right)\ge (x+y+z)\left(\frac{9}{2(x+y+z)}\right) =\frac{9}{2}$$By multiplying all the above three inequalities, we get $$(x + y+ z+ 3)\left(\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}\right)(x+y+z)^3\ge 6 \cdot \frac{9}{2}\cdot(\sqrt{x}+\sqrt{y}+\sqrt{z})^2=27(\sqrt{x}+\sqrt{y}+\sqrt{z})^2$$Just notice that $$\frac{x + 3}{y + z} + \frac{y + 3}{z + x} + \frac{z + 3}{x + y} + 3=(x+y+z+3)\left(\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}\right)$$Then, we are finished.

This has an ugly general one...

Generalization 1 Let $x,y,z$ be positive reals such that $xy+yz+zx=p$. Then prove that $$\dfrac{x^k+\lambda }{y+z}+\dfrac{y^k+\lambda }{z+x}+\dfrac{z^k+\lambda }{x+y}+p\geq \dfrac{\sqrt{\left(3p\right)^k}\left(\lambda .3^k +p\sqrt{p}\left(\sqrt{3^kp^{k-3}}+2\sqrt{3^{2k-3}}\right)\right)\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2}{2.3^{k-1}\left(x+y+z\right)^{k+2}}$$

We first prove that $x+y+z\ge 3$. Suppose, FTSOC, that this is not the case, and there exist $x,y,z$ such that $x+y+z<3$. Squaring both sides gives $x^2+y^2+z^2+2(xy+yz+zx) = x^2+y^2+z^2+6 < 9$, for $x^2+y^2+z^2<3$. But, by Muirhead, $x^2+y^2+z^2\ge xy+yz+zx=3$, contradiction. Thus proved. Now, rewrite the original inequality as \[(x+y+z+3)\sum_{cyc} \frac{1}{y+z} \ge \frac{(\sqrt{x}+\sqrt{y}+\sqrt{z})^2}{(x+y+z)^3}.\]Note that by Titu, we have that both \begin{align*} \sum_{cyc}\frac{1}{y+z} &\ge \frac{9}{2(x+y+z)} \\ \frac{3}{x+y+z} = \sum_{cyc} \frac{x}{\tfrac{1}{3}(x+y+z)^2} &\ge \frac{(\sqrt{x}+\sqrt{y}+\sqrt{z})^2}{(x+y+z)^2}. \end{align*}Thus, it suffices to prove that \[\frac{9(x+y+z+3)}{2(x+y+z)} \ge 27\cdot \frac{3}{(x+y+z)^2}.\]Let $s = x+y+z$. We can then rearrange this inequality as $s^2+3s-18 = (s+6)(s-3) \ge 0$, which is true for all $s\ge 3$, and hence we are done. $\blacksquare$