Example 6.1.2. Let \(x, y, z, t\) be real numbers satisfying \(x y + y z + z t + t x = 1\). Find the minimum of the expression
\[5 x^{2} + 4 y^{2} + 5 z^{2} + t^{2}\]
Solution. We choose a positive number $l<5$ and apply AM-GM inequality
$$\begin{aligned}
l x^{2}+2 y^{2} & \geq 2 \sqrt{2 l} x y \\
2 y^{2}+l z^{2} & \geq 2 \sqrt{2 l} y z \\
(5-l) z^{2}+1 / 2 t^{2} & \geq \sqrt{2(5-l)} z t \\
1 / 2 t^{2}+(5-l) x^{2} & \geq \sqrt{2(5-l)} t x
\end{aligned}$$

Summing up these results, we conclude that
$$5 x^{2}+4 y^{2}+5 z^{2}+t^{2} \geq 2 \sqrt{2 l}(x y+t z)+\sqrt{2(5-l)}(z t+t x)$$

The condition $x y+y z+z t+t x=1$ suggests us to choose a number $l(0 \leq l \leq 5)$ such that $2 \sqrt{2 l}=\sqrt{2(5-l)}$. A simple calculation yields $l=1$, thus, the minimum of $5 x^{2}+4 y^{2}+5 z^{2}+t^{2}$ is $2 \sqrt{2}$.