5. let $a$ and $b$ be fixed positive numbers. Depending on $a$ and $b$, find the smallest possible value of the sum

$$
\frac{x^{2}}{(a y+b z)(a z+b y)}+\frac{y^{2}}{(a z+b x)(a x+b z)}+\frac{z^{2}}{(a x+b y)(a y+b x)}
$$

where $x, y, z$ are positive real numbers.

## Solution
Be in the following

$$
A=\frac{x^{2}}{(a y+b z)(a z+b y)}+\frac{y^{2}}{(a z+b x)(a x+b z)}+\frac{z^{2}}{(a x+b y)(a y+b x)}
$$

The assumption that the minimum of $A$ is assumed for $x=y=z$ is quite obvious. Substituting $x=y=z$ yields the value $3 /(a+b)^{2}$. We now give three proofs for the inequality

$$
A \geq \frac{3}{(a+b)^{2}}
$$

## 1st proof

The following applies with the help of AM-GM

$$
\begin{aligned}
(a x+b y)(a y+b x) & =x y\left(a^{2}+b^{2}\right)+\left(x^{2}+y^{2}\right) a b \\
& \leq \frac{1}{2}\left(x^{2}+y^{2}\right)\left(a^{2}+b^{2}\right)+\left(x^{2}+y^{2}\right) a b=\frac{(a+b)^{2}}{2}\left(x^{2}+y^{2}\right)
\end{aligned}
$$

These and the analogous estimates provide

$$
A \geq \frac{2}{(a+b)^{2}}\left(\frac{x^{2}}{y^{2}+z^{2}}+\frac{y^{2}}{z^{2}+x^{2}}+\frac{z^{2}}{x^{2}+y^{2}}\right)
$$

The large bracket is now always $\geq 3 / 2$, which concludes the proof (Nesbit's inequality). To see this, you can use C.S., for example. On the one hand, the following applies

$$
\left(\frac{u}{v+w}+\frac{v}{w+u}+\frac{w}{u+v}\right)(u(v+w)+v(w+u)+w(u+v)) \geq(u+v+w)^{2}
$$

but on the other hand

$u(v+w)+v(w+u)+w(u+v) \leq \frac{2}{3}\left(u^{2}+v^{2}+w^{2}+2 u v+2 v w+2 w u\right)=\frac{2}{3}(u+v+w)^{2}$,

Combining the two inequalities produces the desired result.

## 2nd proof

We can assume oBdA $x \geq y \geq z$. Then $x^{2} \geq y^{2} \geq z^{2}$ and

$$
(a x+b y)(a y+b x) \geq(a z+b x)(a x+b z) \geq(a y+b z)(a z+b y)
$$

With the help of Tchebychef and AM-HM, the following now follows

$$
\begin{aligned}
A & \geq \frac{\left(x^{2}+y^{2}+z^{2}\right)}{3}\left(\frac{1}{(a y+b z)(a z+b y)}+\frac{1}{(a z+b x)(a x+b z)}+\frac{1}{(a x+b y)(a y+b x)}\right) \\
& \geq 3\left(x^{2}+y^{2}+z^{2}\right) \frac{1}{(a y+b z)(a z+b y)+(a z+b x)(a x+b z)+(a x+b y)(a y+b x)}
\end{aligned}
$$

Furthermore

$$
\begin{aligned}
& (a y+b z)(a z+b y)+(a z+b x)(a x+b z)+(a x+b y)(a y+b x) \\
= & \left(a^{2}+b^{2}\right)(x y+y z+z x)+2 a b\left(x^{2}+y^{2}+z^{2}\right) \\
\leq & \left(a^{2}+b^{2}\right)\left(x^{2}+y^{2}+z^{2}\right)+2 a b\left(x^{2}+y^{2}+z^{2}\right)=(a+b)^{2}\left(x^{2}+y^{2}+z^{2}\right)
\end{aligned}
$$

All in all, as desired

$$
A \geq 3 \frac{x^{2}+y^{2}+z^{2}}{(a+b)^{2}\left(x^{2}+y^{2}+z^{2}\right)}=\frac{3}{(a+b)^{2}}
$$

## 3rd proof

According to C.S.

$$
\begin{gathered}
\left(x^{2}(a y+b z)(a z+b y)+y^{2}(a z+b x)(a x+b z)+z^{2}(a x+b y)(a y+b x)\right) \times A \\
\geq\left(x^{2}+y^{2}+z^{2}\right)^{2}
\end{gathered}
$$

Now complete the solution in a similar way to the 2nd proof.