Find all strictly positive real numbers $c$ such that for all positive real numbers $x, y, z$ we have:

$$
\frac{x^{4}}{\left(y^{2}+1\right)\left(z^{2}+1\right)}+\frac{y^{4}}{\left(x^{2}+1\right)\left(z^{2}+1\right)}+\frac{z^{4}}{\left(x^{2}+1\right)\left(y^{2}+1\right)}+\frac{6}{1+c(x \sqrt{x}+y \sqrt{y}+z \sqrt{z})^{4 / 3}}>3
$$

## Solutions
First, when we see this kind of exercise where we need to determine a constant, it's always a good idea to look at the extreme cases. Here, it involves the case $x=y=z$, some exercises are with the case $x>0, y=z=0$. For lack of a better idea, it's a good strategy to look at these two cases. I skip the case $y=z=0$, since here it does not provide good information. Let's move on to the case $x=y=z$, the inequality becomes:

$$
\frac{3 x^{4}}{\left(x^{2}+1\right)^{2}}+\frac{6}{1+3^{4 / 3} c x^{2}}-3>0
$$

To simplify the notation, I will write $d=3^{4 / 3} c$. We simplify the left term by 3 and put everything over a common denominator:

$$
\frac{x^{4}\left(d x^{2}+1\right)+2\left(x^{2}+1\right)^{2}-\left(x^{2}+1\right)^{2}\left(d x^{2}+1\right)}{\left(x^{2}+1\right)^{2}\left(d x^{2}+1\right)}
$$

The denominator is always positive and does not affect the sign of the fraction, so we can drop it and focus on the numerator:

$$
d x^{6}+x^{4}+2 x^{4}+4 x^{2}+2-\left(d x^{6}+(2+d) x^{4}+(2 d) x^{2}+1\right)=(1-d) x^{4}+(2-2 d) x^{2}+1
$$

We see that if $d>1$, then the leading coefficient is negative, so for sufficiently large $x$ this expression is negative, and if $d \leq 1$ (i.e., $c \leq 3^{-4 / 3}$) then the expression is always strictly positive. Now we have a good idea of the answer and a necessary condition.

For the next part, recall the following notation:

$$
M(p)=\left(\frac{x^{p}+y^{p}+z^{p}}{3}\right)^{1 / p}
$$

the mean of order $p$.

Let's move on to the sufficient condition. Rather than manipulating the ugly term full of $3/2$ powers, we focus on the three symmetric terms. We can assume that $x \geq y \geq z$ and apply Chebyshev's inequality with the two sequences $x^{4} \geq y^{4} \geq z^{4}$ and $\frac{1}{\left(y^{2}+1\right)\left(z^{2}+1\right)} \geq \frac{1}{\left(x^{2}+1\right)\left(z^{2}+1\right)} \geq$ $\frac{1}{\left(x^{2}+1\right)\left(y^{2}+1\right)}$:

$$
\begin{gathered}
(*)=\frac{x^{4}}{\left(y^{2}+1\right)\left(z^{2}+1\right)}+\frac{y^{4}}{\left(x^{2}+1\right)\left(z^{2}+1\right)}+\frac{z^{4}}{\left(x^{2}+1\right)\left(y^{2}+1\right)} \\
\leq \frac{1}{3}\left(x^{4}+y^{4}+z^{4}\right)\left(\frac{1}{\left(y^{2}+1\right)\left(z^{2}+1\right)}+\frac{1}{\left(x^{2}+1\right)\left(z^{2}+1\right)}+\frac{1}{\left(x^{2}+1\right)\left(y^{2}+1\right)}\right)
\end{gathered}
$$

Let's start with the term with the fractions. It is quite easy to put everything over a common denominator, so let's tackle it:

$$
\frac{1}{\left(y^{2}+1\right)\left(z^{2}+1\right)}+\frac{1}{\left(x^{2}+1\right)\left(z^{2}+1\right)}+\frac{1}{\left(x^{2}+1\right)\left(y^{2}+1\right)}=\frac{3+x^{2}+y^{2}+z^{2}}{\left(x^{2}+1\right)\left(y^{2}+1\right)\left(z^{2}+1\right)}
$$

We then use the arithmetic-geometric mean inequality on the denominator:

$$
\left(x^{2}+1\right)\left(y^{2}+1\right)\left(z^{2}+1\right) \leq\left(\frac{3+x^{2}+y^{2}+z^{2}}{3}\right)^{3}=\left(1+M(2)^{2}\right)^{3}
$$

One of the terms $\left(1+M(2)^{2}\right)$ cancels out with the numerator, and what remains is

$$
\frac{1}{\left(y^{2}+1\right)\left(z^{2}+1\right)}+\frac{1}{\left(x^{2}+1\right)\left(z^{2}+1\right)}+\frac{1}{\left(x^{2}+1\right)\left(y^{2}+1\right)} \geq \frac{3}{\left(1+M(2)^{2}\right)^{2}}
$$

We then tackle the other term:

$$
\frac{1}{3}\left(x^{4}+y^{4}+z^{4}\right)=M(4)^{4} \geq M(2)^{4}
$$

by the generalized mean inequality.

We can now combine the two terms and we get:

$$
(*) \geq \frac{3 M(2)^{4}}{\left(M(2)^{2}+1\right)^{2}}=3-\frac{6}{\left(M(2)^{2}+1\right)}+\frac{1}{(M(2)+1)^{2}}
$$

So, by moving $\frac{6}{\left(M(2)^{2}+1\right)}$ to the other side, and using $M(2) \geq M(3 / 2)$:

$$
(*)+\frac{6}{\left(M(3 / 2)^{2}+1\right)} \geq(*)+\frac{6}{\left(M(2)^{2}+1\right)}>3
$$

and the term on the left is the term from the statement with $c=3^{-4 / 3}$, so we have the sufficient condition.

Finally, since the left term is decreasing in $c$, this means that all values less than $3^{-4 / 3}$ also work. The correct answer is:

$$
\left.c \in] 0,3^{-4 / 3}\right]
$$

## 3 Geometry Test (4h)

## Statements