Inequality between altitude and sides in triangle Let $a,b,c$ be the side lengths and $h_a,h_b,h_c$ the altitudes each connect a vertex to the opposite side and are perpendicular to that side. Then we need to prove $h_a^2+h_b^2+h_c^2\leq\dfrac14(a+b+c)^2$. 
I know the inequality $h_a^2+h_b^2+h_c^2\leq\dfrac34(a^2+b^2+c^2)$ by using Cauchy inequality. But I could not extend the proof for the above inequality. Does one help me to prove this? 
 A: Using 2(Area) = $r(a+b+c) = ah_a = bh_b = ch_c$ the inequality can be written in the more appealing form
$$ \sum \frac {1}{a^2} \leq \frac{1}{r^2} $$
which is the question of maximizing $\sum a^{-2}$ for fixed $r$. 
Maybe that has a geometric solution?  Of course there is equality for an equilateral triangle.
Descartes formula seems potentially useful:
https://en.wikipedia.org/wiki/Descartes%27_theorem
A: Following ASCII advocate's idea, the problem boils down to proving that:
$$ 4\Delta^2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\leq (a+b+c)^2 \tag{1}$$
or, using Heron's formula,
$$ (a+b-c)(a-b+c)(-a+b+c)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\leq (a+b+c). \tag{2}$$
Through Ravi substitution that turns out to be equivalent to:
$$ 8xyz\left(\frac{1}{(x+y)^2}+\frac{1}{(x+z)^2}+\frac{1}{(y+z)^2}\right)\leq 2(x+y+z)\tag{3} $$
or to:
$$ \sum_{cyc}\frac{1}{(x+y)^2}\leq\sum_{cyc}\frac{1}{4xy}\tag{4}$$
that is trivial as a consequence of the AM-GM inequality.
A: for known one the proof is
$$h_a^2\le m_a^2=\dfrac {2b^2+2c^2-a^2} 4$$
$$h_b^2\le m_b^2=\dfrac {2a^2+2c^2-b^2} 4$$
$$h_c^2\le m_c^2=\dfrac {2b^2+2a^2-c^2} 4$$
adding these three we get the desired result
A: $h_a^2\le \omega_a^2=\dfrac {4s(s-a)bc} {(b+c)^2}\le s(s-a)$
and similarly
$h_b^2\le s(s-b),\ $
$h_c^2\le s(s-c)$
adding them
$$h_a^2 +h_b^2+h_c^2\le s^2=\dfrac 1 4 (a+b+c)^2$$
