IRL_ABooklet_2023

By the triangle inequality, $a<b+c$, hence $a<2b$, using $b=c$. Therefore $a^2<4bc$, as required.

If $x>1$, and $a=x^2+1$, $b=2x$, $c=x^2-1$, then $$b^2+c^2=4x^2+(x^4-2x^2+1)=x^4+2x^2+1=a^2,$$ and so $a$, $b$, $c$ are side lengths of a right-angled triangle with hypotenuse $a$. Then $$\begin{array}{rl}{a}^2-4bc& =(x^2+1{)}^2-8x(x^2-1)\\ & =x^4-8x^3+2x^2+8x+1\\ & =x^3(x-8)+2x^2+8x+1.\end{array}$$ Clearly, the polynomial $x^3(x-8)+2x^2+8x+1$ is positive if $x$ is large enough, e.g., any $x\ge 8$ will do. So, as long as $x\ge 8$ we have a right-angled triangle in which $a^2>4bc$.

Alternatively, we may find such examples as follows. If $a$ is the length of the hypotenuse of the right-angled triangle $ABC$, its area ($ABC$) is equal to $bc/2$. Hence, the desired inequality $a^2>4bc$ is equivalent to $a^2>8(ABC)$. If $h$ is the height of triangle $ABC$ on the hypotenuse $a$, then ($ABC$) = $ah/2$ and the desired inequality becomes $a>4h$. In the diagram below, this means that $A$ should be on the semicircle but not above the dashed line. To make this more explicit, let the foot of the altitude from $A$ divide the hypotenuse into segments of length $p$ and $q$, and recall that $h^2=pq=p(a-p)$. Thus, we want $a^2>16h^2=16p(a-p)$, i.e. $a^2-16pa+16p^2>0$. This is equivalent to $${\left(\frac{a}{p}-8\right)}^2>48,\text{ i.e. }\left|\frac{a}{p}-8\right|>4\sqrt{3}.$$ As $0<p<a$ in our case, given $a$ we can choose any $p$ such that $$0<p<\frac{2-\sqrt{3}}{4}\text{or}\frac{2+\sqrt{3}}{4}<p<a.$$ Because $c^2=p^2+h^2=p(p+q)=ap$ and $b^2=aq=a(a-p)$, the value of $p$ determines $b=\sqrt{a^2-ap}$ and $c=\sqrt{ap}$ for any given $a$. For example, with $a=41$ and $p=81/41$, we obtain $b=40$ and $c=9$. In this example, $a^2=1681>1440=4bc$.