This problem was proposed by me.

Kimchiks926 wrote: The positive real numbers $x,y,z$ satisfy $xy+yz+zx=1$. Prove that: $$ 2(x^2+y^2+z^2)+\frac{4}{3}(\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}) \ge 5 $$ The following inequality a bit of stronger. Let $x$, $y$ and $z$ be positive numbers such that $xy+xz+yz=1$. Prove that: $$x^2+y^2+z^2+4\left(\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}\right)\geq10.$$

(Almost) one-liner: set $t=x^2+y^2+z^2$, then $t\ge 1$. We have by AM-HM inequality that \[ 2(x^2+y^2+z^2)+\frac{4}{3}(\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}) \ge 2t+\frac{12}{t+3}. \]Finally, $2t+12/(t+3)\ge 5$ as it is equivalent to $(t-1)(2t+3)\ge 0$ which is clear for $t\ge 1$.

Straightforward by Titu's lemma (applied on $\frac{1}{x^2+1}$ and the other two) and $x^2 + y^2 + z^2 \geq xy + yz + zx$.

Alternatively, just add up the inequalities \[2x^2+\frac{4}{3(x^2+1)} \ge \frac{5}{4}\left(x^2-\frac{1}{3}\right)+\frac{5}{3}\]which is equivalent to $\left(x^2-\frac{1}{3}\right)^2 \ge 0$ and hence true, and finally use that $x^2+y^2+z^2 \ge 1$.

arqady wrote: Kimchiks926 wrote: The positive real numbers $x,y,z$ satisfy $xy+yz+zx=1$. Prove that: $$ 2(x^2+y^2+z^2)+\frac{4}{3}(\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}) \ge 5 $$ The following inequality a bit of stronger. Let $x$, $y$ and $z$ be positive numbers such that $xy+xz+yz=1$. Prove that: $$x^2+y^2+z^2+4\left(\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}\right)\geq10.$$ Stronger, let $x$, $y$ and $z$ be positive numbers such that $xy+xz+yz=1$. Prove that: $$x^2+y^2+z^2+4\left(\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}\right)\geq10+\frac{(x-y)^2(y-z)^2(z-x)^2}{(xy^2+yz^2+zx^2)(x^2y+y^2z+z^2x)}.$$

no_room_for_error wrote: Stronger, let $x$, $y$ and $z$ be positive numbers such that $xy+xz+yz=1$. Prove that: $$x^2+y^2+z^2+4\left(\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}\right)\geq10+\frac{(x-y)^2(y-z)^2(z-x)^2}{(xy^2+yz^2+zx^2)(x^2y+y^2z+z^2x)}.$$ A bit of stronger: Let $x$, $y$ and $z$ be positive numbers such that $xy+xz+yz=1$. Prove that: $$x^2+y^2+z^2+4\left(\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}\right)\geq10+(x-y)^2(y-z)^2(z-x)^2\left(\frac{1}{(xy^2+yz^2+zx^2)(x^2y+y^2z+z^2x)}+\frac{1569x^2y^2z^2}{(x+y)^4(x+z)^4(y+z)^4}\right).$$

Note that $\frac{1}{x^2+1}=\frac{1}{(x+y)(x+z)}$. By AM-GM, $$\sqrt{\frac{3}{4}}(x+y)+\sqrt{\frac{3}{4}}(x+z)+\frac{4}{3}\frac{1}{(x+y)(x+z)}\ge 3.$$Futhermore, $(x+y+z)^2\ge 3(xy+yz+zx)=3\implies x+y+z\ge \sqrt{3}$, which means $$x^2+y^2+z^2+2=x^2+y^2+z^2+2(xy+yz+zx)=(x+y+z)^2\ge\sqrt{3}(x+y+z).$$By summing the two inequalities (the first one cyclicly), we are done.

Beautiful

Claim 1: $x+y+z\geq \sqrt{3}$. Let $s=x+y+z$. We then have $$s^2=x^2+y^2+z^2+2\geq3,$$so $s\geq \sqrt{3}.$ Claim 2: The function $$f(x)=2x^2+\frac{4}{3}\frac{1}{x^2+1}$$is increasing and convex for positive $x$. This can be seen by taking derivatives twice. Since $f$ is convex and increasing, we have $$f(x)+f(y)+f(z)\geq 3f(\frac{x+y+z}{3})\geq 3f(\frac{\sqrt{3}}{3})=5,$$so we are done.

We have $A=x^2+y^2+z^2$ $\ge$ $1$ $\cfrac{1}{x^2+1}$$+$$\cfrac{1}{y^2+1}$$+$$\cfrac{1}{z^2+1}$ $\ge$ $\cfrac{9}{A+3}$ $2A+ \cfrac{12}{A+3}=\cfrac{A+3}{2}+\cfrac{A+3}{2}+\cfrac{8}{A+3}+\cfrac{8}{A+3}+A-\cfrac{4}{A+3}-3$ By AM-GM $\cfrac{A+3}{2}+\cfrac{A+3}{2}+\cfrac{8}{A+3}+\cfrac{8}{A+3} \ge 8$ Also we know that $A \ge 1 \ge \cfrac{4}{A+3}$ So $2A+ \cfrac{12}{A+3} \ge 8-3=5$

Let $q=yz+zx+xy=1$ and $k=\dfrac{x^2+y^2+z^2}{q}.$ Now, homogenize the given inequality as $$\frac{2(x^2+y^2+z^2)}{q}+\frac 43\sum_{cyc}\frac{q}{x^2+q}.$$Now, use $\textcolor{red}{\textit{Titu's Lemma,}}$ to obtain $$\sum_{cyc}\frac{q}{x^2+q}\ge \frac{9q}{x^2+y^2+z^2+3q}=\frac{9}{k+3}.$$Therefore, we need to prove that $$2k+\frac{12}{k+3}\ge 5\iff \boxed{(2k+3)(k-1)\ge 0.}$$Which is clearly true since $k\ge 1.$ Equality holds iff $x=y=z=\dfrac{1}{\sqrt 3}.$ $\color{green}{\cal{QED.}}$

Let $x^2+y^2+z^2=t$ and $t\geq 1$. Use Bergström's Inequality $$LHS\geq 2t+\dfrac{12}{t+3}\geq 5\Longleftrightarrow 2t^2+6t+12\geq 5t+15$$which is true for $t\geq 1$.