3. Determine the smallest and largest value of the function

$$
f(x, y, z, t)=\frac{a x^{2}+b y^{2}}{a x+b y}+\frac{a z^{2}+b t^{2}}{a z+b t},(a>0, b>0)
$$

under the conditions $x+z=y+t=1, x, y, z, t \geq 0$.
Solution. From $0 \leq x \leq 1, 0 \leq y \leq 1$ it follows that $x^{2} \leq x, y^{2} \leq y$, and therefore

$$
a x^{2}+b y^{2} \leq a x+b y
$$

Similarly, it can be shown that

$$
a z^{2}+b t^{2} \leq a z+b t
$$

therefore

$$
f(x, y, z, t)=\frac{a x^{2}+b y^{2}}{a x+b y}+\frac{a z^{2}+b t^{2}}{a z+b t} \leq 1+1=2
$$

Since $f(1,0,0,1)=2$, we conclude that 2 is the maximum value of the function $f$ under the given conditions.

We will prove that under the given conditions, the following holds

$$
\frac{a x^{2}+b y^{2}}{a x+b y} \geq \frac{a x+b y}{a+b}
$$

Indeed, the last inequality is equivalent to the inequality

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

or to the inequality

$$
a b(x-y)^{2} \geq 0
$$

which is clearly true. Therefore

$$
f(x, y, z, t)=\frac{a x^{2}+b y^{2}}{a x+b y}+\frac{a z^{2}+b t^{2}}{a z+b t} \geq \frac{a x+b y}{a+b}+\frac{a z+b t}{a+b}=\frac{a(x+z)+b(y+t)}{a+b}=1
$$

Since $f\left(, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)=1$, we conclude that 1 is the minimum value of the function $f$ under the given conditions.