6. Find all positive real numbers $p$ such that the inequality

$$
\sqrt{a^{2}+p b^{2}}+\sqrt{b^{2}+p a^{2}} \geqq a+b+(p-1) \sqrt{a b}
$$

holds for any pair of positive real numbers $a, b$.

(Jaromír Šimša)
Solution. For the pair $a=b=1$, we get the inequality for the parameter $p>0$, which we solve:

$$
\begin{aligned}
2 \sqrt{p+1} & \geqq p+1 \\
2 & \geqq \sqrt{p+1} \\
p & \leqq 3
\end{aligned}
$$

We will now show that the required property holds for every $p \in(0,3\rangle$.

For $p \in(0,1\rangle$, the given inequality is trivially satisfied, since

$$
\sqrt{a^{2}+p b^{2}}>a, \quad \sqrt{b^{2}+p a^{2}}>b \quad \text { and } \quad(p-1) \sqrt{a b} \leqq 0 .
$$

Therefore, we will focus on the case $p \in(1,3\rangle$. The left-hand side $L$ of the inequality to be proven can be interpreted as the magnitude of two vectors $(a, b \sqrt{p}),(b, a \sqrt{p}) \in \mathbb{R}^{2}$, so by the triangle inequality,

$$
\begin{aligned}
L=\sqrt{a^{2}+p b^{2}}+\sqrt{b^{2}+p a^{2}} & =|(a, b \sqrt{p})|+|(b, a \sqrt{p})| \geqq \\
& \geqq|(a+b,(a+b) \sqrt{p})|=(a+b) \sqrt{1+p} .
\end{aligned}
$$

For the right-hand side $P$, using the inequality between the arithmetic and geometric means, we obtain an upper bound

$$
P=a+b+(p-1) \sqrt{a b} \leqq a+b+(p-1) \frac{a+b}{2}=\frac{(p+1)(a+b)}{2} .
$$

The inequality $L \geqq P$ is thus proven, since the stronger inequality

$$
(a+b) \sqrt{p+1} \geqq \frac{(p+1)(a+b)}{2}
$$

is equivalent to the inequality $\sqrt{p+1} \leqq 2$, which is clearly satisfied for every $p \in(1,3\rangle$.

Note. The estimate (1) can also be obtained by applying the Cauchy-Schwarz inequality to the pairs $(a, b \sqrt{p})$ and $(1, \sqrt{p})$: from the inequality

$$
a+p b \leqq \sqrt{a^{2}+p b^{2}} \cdot \sqrt{1+p}
$$

we derive the first of the inequalities

$$
\sqrt{a^{2}+p b^{2}} \geqq \frac{a+p b}{\sqrt{1+p}}, \quad \sqrt{b^{2}+p a^{2}} \geqq \frac{b+p a}{\sqrt{1+p}}
$$

the second is derived analogously.