Right triangle, feet of altitude, angle bisector and median

[maxkor]

Let $\triangle ABC$ be a right-angled triangle with $\angle A = 90^{\circ}$, and $AB < AC$. Let points $D, E, F$ be located on side $BC$ such that $AD$ is the altitude, $AE$ is the internal angle bisector, and $AF$ is the median.

Prove that $3AD + AF > 4AE$

My solution. Can you check it is right? (* 3 times I've used inequality AM,GM)

 

[jvicens]

Yes, your solution is correct! Here is a detailed explanation of your proof:

Let $a = BC$, $b = AC$, and $c = AB$. Since $\angle A = 90^{\circ}$, we know that $c$ is the hypotenuse of $\triangle ABC$. Since $AB < AC$, we have $c > b$, which means that $a^2 + b^2 > b^2 + c^2$. This can also be rewritten as $a^2 > c^2$, which implies $a > c$.

First, let's find the lengths of $AD$, $AE$, and $AF$ in terms of $a$, $b$, and $c$.

Since $AD$ is the altitude, we know that $AD = \frac{bc}{a}$.

To find $AE$, we can use the angle bisector theorem, which states that the length of the angle bisector of a triangle is equal to the ratio of the sides it divides, multiplied by the length of the side opposite the angle bisector. In this case, we have $\frac{AE}{CE} = \frac{AB}{BC} = \frac{c}{a}$. Since $CE = a - AE$, we can rewrite this as $\frac{AE}{a - AE} = \frac{c}{a}$. Cross-multiplying, we get $aAE = c(a - AE)$, which simplifies to $AE = \frac{ac}{a + c}$.

Lastly, we can use the fact that $AF$ is the median to find its length. Since $AF$ divides $BC$ into two equal parts, we have $AF = \frac{a}{2}$.

Now, let's substitute these values into the inequality $3AD + AF > 4AE$:

$3\left(\frac{bc}{a}\right) + \frac{a}{2} > 4\left(\frac{ac}{a + c}\right)$

Multiplying both sides by $a + c$, we get:

$3bc + \frac{a(a + c)}{2} > 4ac$

$6bc + a^2 + ac > 8ac$

$a^2 + 6bc > 7ac$

Since $a > c$, we can multiply both sides by $a$ without changing the inequality