a) Determine the largest real number $A$ with the following property: For all non-negative real numbers $x,y,z$, one has \[\frac{1+yz}{1+x^2}+\frac{1+zx}{1+y^2}+\frac{1+xy}{1+z^2} \ge A.\]b) For this real number $A$, find all triples $(x,y,z)$ of non-negative real numbers for which equality holds in the above inequality.
Problem
Source: Germany 2021, Problem 5
Tags: inequalities, inequalities proposed, Symmetric inequality, equality case
16.06.2021 11:02
$y=z=0,x\to\infty$ gives $A\leq2$. \begin{align*} \frac{1+yz}{1+x^2}+\frac{1+zx}{1+y^2}+\frac{1+xy}{1+z^2}>2\\ \Leftrightarrow\sum_{cyc}(1+yz)\left(1+y^2\right)\left(1+z^2\right)>2\left(1+x^2\right)\left(1+y^2\right)\left(1+z^2\right)\\ \Leftrightarrow\left(x^3y^3+y^3z^3+z^3x^3-3x^2y^2z^2\right)+\left(yz(y-z)^2+zx(z-x)^2+xy(x-y)^2\right)+x^2y^2z^2+x^2y^2+y^2z^2+z^2x^2+xy+yz+zx+1>0 \end{align*}Thus $A=2$. There are no equality cases.
16.06.2021 11:23
pi_quadrat_sechstel wrote: $y=z=0,x\to\infty$ gives $A\leq2$. \begin{align*} \frac{1+yz}{1+x^2}+\frac{1+zx}{1+y^2}+\frac{1+xy}{1+z^2}>2\\ \Leftrightarrow\sum_{cyc}(1+yz)\left(1+y^2\right)\left(1+z^2\right)>2\left(1+x^2\right)\left(1+y^2\right)\left(1+z^2\right)\\ \Leftrightarrow\left(x^3y^3+y^3z^3+z^3x^3-3x^2y^2z^2\right)+\left(yz(y-z)^2+zx(z-x)^2+xy(x-y)^2\right)+x^2y^2z^2+x^2y^2+y^2z^2+z^2x^2+xy+yz+zx+1>0 \end{align*}Thus $A=2$. There are no equality cases. ? When $A=2$, $(x,y,z)=?$
16.06.2021 18:40
sqing wrote: When $A=2$, $(x,y,z)=?$ I don't understand your objection. It was proved that $A=2$ is the largest constant for which the inequality is true. It was also proved that there is no equality case for $A=2$. So the answer to part b) is that there are no such triples. Why does this mean that the problem is wrong?
14.07.2021 04:50
semms like xz can be anything if z=0 x to infinity maybe z=y=t,x=t^(-1/2), still doesn't change the fact that no triples fit
17.07.2021 02:39
WLOG $x\ge y,z$ then $$LHS > \frac{1+zx}{1+y^2}+\frac{1+xy}{1+z^2}\ge \frac{1+z^2}{1+y^2}+\frac{1+y^2}{1+z^2}\ge 2$$ By AM-GM.
26.05.2023 23:04
The answer is $A=2$, which can be achieved by letting $x=y=0$ and $z\to\infty$. Let the sum be $T$ Assuming wlog that $z\ge x,y$, we have that \[\sum_{\mathrm{cyc}}\frac{1+yz}{1+x^2}\ge \frac{1+x^2}{1+y^2}+\frac{1+y^2}{1+x^2}\ge 2\]by AM-GM. For equality to hold, we must have $(1+x^2)^2=(1+y^2)^2\Longleftrightarrow x=y$ as $x,y\ge 0$. Putting that into the original inequality, we see that \[T\ge 2\frac{1+xz}{1+x^2}+\frac{1+x^2}{1+z^2}>2\frac{1+xz}{1+x^2}\ge 2\]so equality can never be reached.