$C=16$ It's just $$(y+z)^2(y+x)^2 \geq 16y^2xz$$which is obvious. Equality holds when $x=y=z$

sqing wrote: Let $\alpha$ be an arbitrary positive real number. Determine for this number $\alpha$ the greatest real number $C$ such that the inequality$$\left(1+\frac{\alpha}{x^2}\right)\left(1+\frac{\alpha}{y^2}\right)\left(1+\frac{\alpha}{z^2}\right)\geq C\left(\frac{x}{z}+\frac{z}{x}+2\right)$$is valid for all positive real numbers $x, y$ and $z$ satisfying $xy + yz + zx =\alpha.$ When does equality occur? (Proposed by Walther Janous) If $x=y=z$ we get $C\le 16$ Then we have to prove inequality when $C=16$, we get $(x+y)^2(y+z)^2(z+x)^2 \ge 16xy^2z(x+z)^2$ $\iff (x+y)^2(y+z)^2 \ge 16 xy^2z $ wich true by AM-GM

Austrian Mathematical Olympiad 2016: Determine the largest constant $C $ such that $(x_1 + x_2 +\cdots+ x_6)^2\geq C \left(x_1(x_2 + x_3) + x_2(x_3 + x_4)+\cdots+ x_6(x_1 + x_2)\right)$ holds for all real numbers $x_1, x_2, \cdots, x_6.$ For this $C,$ determine all $x_1, x_2, \cdots, x_6 $ such that equality holds. (Proposed by Walther Janous)

sqing wrote: Austrian Mathematical Olympiad 2016: Determine the largest constant $C $ such that $(x_1 + x_2 +\cdots+ x_6)^2\geq C \left(x_1(x_2 + x_3) + x_2(x_3 + x_4)+\cdots+ x_6(x_1 + x_2)\right)$ holds for all real numbers $x_1, x_2, \cdots, x_6.$ For this $C,$ determine all $x_1, x_2, \cdots, x_6 $ such that equality holds. (Proposed by Walther Janous) Setting $x_1=x_2= \ldots=x_6$ yields $C \leqslant 3$. So, we prove the given inequality for $C=3$. We notice firstly that $x_1(x_2+x_3) +\ldots+x_6(x_1+x_2)=(x_1+x_4)(x_2+x_5)+(x_3+x_6)(x_1+x_4)+(x_2+x_5)(x_3+x_6).$ We set $x_1+x_4=m, x_2+x_5=n, x_3+x_6=p$, and we need to prove that $(m+n+p)^2 \geqslant 3(mn+np+pm)$, which is obvious. Equality holds if and only if $x_1+x_4=x_2+x_5=x_3+x_6$.

Supercali wrote: $C=16$ It's just $$(y+z)^2(y+x)^2 \geq 16y^2xz$$which is obvious. Equality holds when $a=b=c$ How do you prove that $C=16$ is the largest possible constant?

Quantum_fluctuations wrote: Supercali wrote: $C=16$ It's just $$(y+z)^2(y+x)^2 \geq 16y^2xz$$which is obvious. Equality holds when $a=b=c$ How do you prove that $C=16$ is the largest possible constant? Because if we put $x=y=z$ we get $C\leq 16$