Inequality between altitude and sides in triangle

Question

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?

Answer 1

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

Answer 2

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

Answer 3

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.

Answer 4

$h_a^2\le \omega_a^2=\dfrac {4s(s-a)bc} {(b+c)^2}\le s(s-a)$

and similarly

$h_b^2\le s(s-b),\ $ $h_c^2\le s(s-c)$

adding them $$h_a^2 +h_b^2+h_c^2\le s^2=\dfrac 1 4 (a+b+c)^2$$