# Proving geometrical inequalities

Let $$u$$ and $$v$$ be the distances of an arbitrary point on the side $$AB$$ of an acute-angled triangle $$ABC$$ to its sides $$BC$$ and $$AC$$. Let $$h_a$$ and $$h_b$$ be the lengths of the altitudes from its vertices $$A$$ and $$B$$ respectively. Prove that  $$min$$ {$$h_a,h_b$$} $$\leq$$ $$u+v$$ $$\leq$$ $$max$$ {$$h_a,h_b$$}  I was able to prove the first part of the question like this: Let the arbitrary point be X. WLOG, let $$h_a\geq h_b$$. Then$$\frac{AX}{XB}={h_a\over v}={u\over h_b}$$ $$h_ah_b=uv$$ $$u+v\geq2\sqrt{uv}=2\sqrt{h_ah_b}\geq2h_b$$ $$Or, u+v\geq min\{h_a,h_b\}$$ But now I'm stuck in the second part.  Edit: The above proof is INCORRECT; the correct proof is given below in @player3236 's answer.

Your idea (of using similar triangles) is correct but the formula is wrong. We should have:

$$\frac {AX}{AB} = \frac {v}{h_b}, \quad \frac {BX}{AB} = \frac {u}{h_a}$$

Hence

$$u+v = h_a\frac {BX}{AB} + h_b\frac {AX}{AB}$$

Therefore

$$\min\{h_a,h_b\}\frac {BX}{AB} + \min\{h_a,h_b\}\frac {AX}{AB} \le h_a\frac {BX}{AB} + h_b\frac {AX}{AB} \le \max\{h_a,h_b\}\frac {BX}{AB} + \max\{h_a,h_b\}\frac {AX}{AB}$$

which simplifies to

$$\min\{h_a,h_b\} \le u+v \le \max\{h_a,h_b\}$$