mihajlon wrote: Let $x$, $y$, $z$ be arbitrary positive numbers such that $xy+yz+zx=x+y+z$. Prove that $$\frac{1}{x^2+y+1} + \frac{1}{y^2+y+1} + \frac{1}{z^2+x+1} \leq 1$$. When does equality occur? Proposed by Marko Radovanovic Do you mean $$\frac{1}{x^2+y+1} + \frac{1}{y^2+z+1} + \frac{1}{z^2+x+1} \leq 1?$$

Yes, Thanks YESMAths

mihajlon wrote: Let $x$, $y$, $z$ be arbitrary positive numbers such that $xy+yz+zx=x+y+z$. Prove that $$\frac{1}{x^2+y+1} + \frac{1}{y^2+z+1} + \frac{1}{z^2+x+1} \leq 1$$. When does equality occur? Proposed by Marko Radovanovic Observe that $(xy+yz+zx)^2=(x+y+z)^2\ge 3(xy+yz+zx)$ which implies that $xy+yz+zx\ge 3$. Then from Holder's Inequality, we have $$\sum_{cyc}\frac{1}{x^2+y+1}=\sum_{cyc}\frac{1+y+z^2}{(x^2+y+1)(1+y+z^2)}\le\sum_{cyc}\frac{1+y+z^2}{(x+y+z)^2}.$$It is thus, sufficient to prove that $(x+y+z)^2\ge \sum_{cyc}(1+y+z^2)=3+(x+y+z)+x^2+y^2+z^2$ which is equivalent to $xy+yz+zx\ge 3$. So, we are done and equality holds only when $x=y=z=1$.

mihajlon wrote: Let $x$, $y$, $z$ be arbitrary positive numbers such that $xy+yz+zx=x+y+z$. Prove that $$\frac{1}{x^2+y+1} + \frac{1}{y^2+z+1} + \frac{1}{z^2+x+1} \leq 1$$. When does equality occur? Proposed by Marko Radovanovic Serbia TST 2009

Its obvious by using a single cauchy shwartz in denominator which implies to prove that sigma x >=3 which is true