
6. Find all real p such that the inequality

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

holds for any real $a$ and $b$.

(Jaromír Šimša)


Solution. If $a=b=1$ the parameter $p>0$ has to satisfy:

$$
\begin{aligned}
2 \sqrt{p+1} & \geqslant p+1 \\
2 & \geqslant \sqrt{p+1} \\
p & \leqslant 3
\end{aligned}
$$

We show that for $p \in(0,3\rangle$ the inequality holds for any real $a$ and $b$. If $p \in(0,1\rangle$ the inequality holds trivially:

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

Let further be $p \in(1,3\rangle$. The left hand side, LHS, of the inequality can be understood as the sum of the lengths of vectors $(a, b \sqrt{p})$ and $(b, a \sqrt{p}) \in \mathbb{R}^{2}$, According to the triangle inequality then

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

For the RHS we have (with the help of AM-GM inequality)

$$
R H S=a+b+(p-1) \sqrt{a b} \leqslant a+b+(p-1) \frac{a+b}{2}=\frac{(p+1)(a+b)}{2} .
$$

Now $L H S \geqslant R H S$ evidently, because even stronger inequality

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

is equivalent to $\sqrt{p+1} \leqslant 2$, which is obviously satisfied for any $p \in(1,3\rangle$.

Remark. We can get (1) using Cauchy-Schwarz inequality for pairs $(a, b \sqrt{p})$ and $(1, \sqrt{p})$ :

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

which implies

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

summing last two inequalities we get (1) and analogously for the second inequality.

The Czech team is supported by the Karel Janeček's foundation.

Účast reprezentačního družstva ČR na 54. mezinárodní matematické olympiádě byla podpořena Nadačním fondem Karla Janečka na podporu vědy a výzkumu, sponzorem matematické olympiády v ČR.

