Exercise 7. Determine the largest real number $a$ and the smallest real number $b$ such that for all non-negative $x, y, z$ we have

$$
a(x+y+z)^{2} \leqslant x^{2}+y^{2}+z^{2}+y z \leqslant b(x+y+z)^{2}
$$
## Solution to Exercise 7

Let's start with $b$. For $x=y=0$ and $z=1$, we observe that if $b$ is valid, we must have $\mathrm{b} \geqslant 1$. Conversely, for all non-negative $x, y, z$, we have

$$
x^{2}+y^{2}+z^{2}+y z \leqslant x^{2}+y^{2}+z^{2}+2 y z+2 z x+2 x y=(x+y+z)^{2}
$$

which proves that $b=1$ is indeed valid. Therefore, the smallest real number $b$ that works is $\mathrm{b}=1$.

For non-negative $x, y, z$, let $f(x, y, z)=x^{2}+y^{2}+z^{2}+y z$. We then observe that

$$
\begin{aligned}
f(x, y, z)-f\left(x, \frac{y+z}{2}, \frac{y+z}{2}\right) & =y^{2}+z^{2}+y z-3\left(\frac{y+z}{2}\right)^{2} \\
& =\frac{1}{4}\left(y^{2}+z^{2}-2 y z\right) \\
& =\frac{1}{4}(y-z)^{2} \\
& \geqslant 0
\end{aligned}
$$

Thus, we have $f(x, y, z) \geqslant f\left(x, \frac{y+z}{2}, \frac{y+z}{2}\right)$. Since $x+y+z=x+\frac{y+z}{2}+\frac{y+z}{2}$, this ensures that, without loss of generality, we can restrict ourselves to triplets of the form $(x, y, y)$, with $x, y$ non-negative. The inequality under study becomes $a(x+2 y)^{2} \leqslant x^{2}+3 y^{2}$. If $x=y=0$, this inequality is clearly true for any $a \geqslant 0$. Therefore, we assume that $(x, y) \neq(0,0)$. Since the inequality is homogeneous, without loss of generality, we can now impose that $x+2 y=1$. It is therefore a matter of finding the largest real number $a$ such that, for all $y \in[0,1]$, we have $a \leqslant(1-2 y)^{2}+3 y^{2}$. Now, we have

$$
(1-2 y)^{2}+3 y^{2}=7 y^{2}-4 y+1=7\left(y-\frac{2}{7}\right)^{2}+\frac{3}{7}
$$

so $(1-2 y)^{2}+3 y^{2} \geqslant \frac{3}{7}$, with equality if and only if $y=\frac{3}{7}$. Finally, the largest real number $a$ that works is $a=\frac{3}{7}$.