Solution during contest: Take $b=a+x$, $c=a+y$ and $d=a+z$ with $x<y<z$ and $x,y,z\in\mathbb{N}^*$. Then the condition $ad=bc$ is equivalent to $a(a+z)=(a+x)(a+y)$, or $az=ax+ay+xy$, which implies $a\mid xy\implies xy=ka$ with $k\in\mathbb{N}^*$. But now $$2a+\sqrt{a}+\sqrt{d}<b+c+1\iff \sqrt{a}+\sqrt{a+z}<x+y+1\iff 2a+z+2\sqrt{a(a+z)}<x^2+y^2+1+2x+2y+2xy.$$Notice that $z=x+y+k$ and from AM-GM (without the possibility of equality) $2\sqrt{a(a+z)}=2\sqrt{(a+x)(a+y)}<2a+x+y$. This implies that $2a+z+2\sqrt{a(a+z)}<4a+2x+2y+k$, meaning that we want to prove $$4a+k\leq x^2+y^2+2xy+1.$$But the latter is obvious, since $x^2+y^2+2xy+1\geq 4xy+1=4ka+1=(4a-1)(k-1)+4a+k\geq 4a+k$.

An approach different than the official solution: Lemma: There exist $x,y,z,t \in \mathbb{N^*}$ such that $a=xy, b=xz, c=yt, d=zt$. The proof is pretty straightforward if we choose $x=(a,b)$, $t=(c,d)$ and rewrite the relation $ad=bc$ in terms of these gcds. Using AM-GM we thus have $$2a+\sqrt{a}+\sqrt{d}=2xy+\sqrt{xy}+\sqrt{zt} < 2xy + \dfrac{x+y+z+t}{2}$$. This inequality is strict, because equality would force $x=y, z=t \Longrightarrow xz=yt \Longrightarrow b=c$, a contradiction. Thus it is sufficient to prove that $$2xy + \dfrac{x+y+z+t}{2} \leq b+c+1 \Longleftrightarrow 2xy + \dfrac{x+y+z+t}{2} \leq xz+yt+1 \Longleftrightarrow 4xy+x+y+z+t \leq 2xz+2yt+2, (*).$$Now notice that $a < b$, so $y < z$. Since $b < c$, we have $xz < yt$, and since $y < z$, it follows that $x < t$. Since $x,y,z,t$ are (strictly) positive integers, we have: $z > y \Rightarrow z-y \geq 1 \Rightarrow z-y-1 \geq 0 \Rightarrow (2x-1)(z-y-1) \geq 0 \Rightarrow 2xz+1 \geq 2xy+2x-y+z, (1)$. Similarly, $t-x-1 \geq 0 \Rightarrow (2y-1)(t-x-1) \geq 0 \Rightarrow 2yt+1 \geq 2xy-x+2y+t, (2)$. Adding $(1)$ and $(2)$ yields $(*)$, so we are done. Remark: If we relax the initial restriction to read $b \leq c$, we can have equality, i.e. $2a+\sqrt{a}+\sqrt{d} = b+c+1$. To find the cases where this does occur, we need to have equality in AM-GM, so $x=y, z=t$, and equality in $(1)$ and $(2)$, so $t-x=z-y=1$. This gives use $x=y=n, z=t=n+1$, where $n \in \mathbb{N^*}$, and thus $$a=n^2, b=c=n(n+1), d=(n+1)^2, $$which do indeed give $2a+\sqrt{a}+\sqrt{d} = b+c+1$.