62nd Czech and Slovak Mathematical Olympiad

If $a=b=1$ the parameter $p>0$ has to satisfy: $$\{\begin{array}{l}2\sqrt{p+1}\ge p+1,\\ 2\ge \sqrt{p+1},\\ p\le 3.\end{array}$$ We show that for $p\in (0,3)$ the inequality holds for any real $a$ and $b$. If $p\in (0,1)$ the inequality holds trivially: $$\sqrt{a^2+p{b}^2}>a,\sqrt{b^2+p{a}^2}>b\text{and}(p-1)\sqrt{ab}\le 0.$$ Let further be $p\in (1,3)$. 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 $$\begin{array}{rl}LHS& =\sqrt{a^2+p{b}^2}+\sqrt{b^2+p{a}^2}=|(a,b\sqrt{p})|+|(b,a\sqrt{p})|\\ & \ge |(a+b,(a+b)\sqrt{p})|=(a+b)\sqrt{1+p}.\end{array}(1)$$ For the RHS we have (with the help of AM-GM inequality) $$\text{RHS}=a+b+(p-1)\sqrt{ab}\le a+b+(p-1)\frac{a+b}{2}=\frac{(p+1)(a+b)}{2}.$$ Now $LHS\ge RHS$ evidently, because even stronger inequality $$(a+b)\sqrt{p+1}\ge \frac{(p+1)(a+b)}{2}$$ is equivalent to $\sqrt{p+1}\le 2$, which is obviously satisfied for any $p\in (1,3)$.