An inequality for a side length of a triangle Let $\Delta ABC$ be a triangle, and $a,b,c$ are the side length $BC$, $CA$, $AB$, respectively. Assume that $c\leq b\leq a$. Let $h$ be the distance from point $A$ to $BC$. Show that $h\leq c\sqrt{1-\frac{c^2}{4b^2}}$.
I have noticed that equality holds when $b=a$, this can be shown by simple property of similar triangles. But I failed to understand how the upper bound holds in general.
Thanks
 A: OK, this turned out to be trickier than I expected.
Let $\beta$ be the angle at point/vertex B. 
One can notice that $\beta \leq \pi /2$ 
(because $b \leq a$). 
So the cosine of $\beta$ is non-negative. 
The given inequality is equivalent to:   
$h_a / c \leq \sqrt{1-\frac{c^2}{4b^2}}$ 
which is the same as: 
$sin(\beta) \leq \sqrt{1-\frac{c^2}{4b^2}}$ 
which is equivalent to 
$sin^2(\beta) \leq 1-\frac{c^2}{4b^2}$ 
which is equivalent to 
$1 - cos^2(\beta) \leq 1-\frac{c^2}{4b^2}$ 
which is equivalent to 
$cos^2(\beta) \geq \frac{c^2}{4b^2}$ 
Now since $cos(\beta)$ is non-negative this is equivalent to:   
$cos(\beta) \geq \frac{c}{2b}$ 
So the last inequality is what we need to prove.  
Let's apply to it the law of cosines:  
$(a^2 + c^2 - b^2)/(2ac) \geq \frac{c}{2b}$ 
After some simple rework, the last one is equivalent to:  
$a^2b + c^2b-b^3-ac^2 \geq 0$ 
i.e. to:  
$(a-b)(ab+b^2-c^2) \geq 0$
The last one is obviously true because $c\leq b\leq a$
and so both $(a-b)$ and $(ab+b^2-c^2)$ are non-negative.  
So by equivalence the inequality we started with is also true.   
When do we reach equality?  
1) when $(a-b)=0$ i.e. when $a=b$ (as you noticed)
2) when $(ab+b^2-c^2) = 0$ but OK... this can never be zero because 
$ab > 0$ and $b^2-c^2 \geq 0$ 
SIDE NOTE: please check carefully that I didn't make any mistake. 
