<s>[INST] True or false? Any triangle we can $h_A^2+h_B^2 \geq 2h_C^2$, where $h_A, h_B, h_C$ are lengths of the heights from vertices A, B and C Let $\triangle ABC$ any triangle and the lengths of the heights from vertices A, B and
C are, respectively,  $h_A$, $h_B$ , $h_C$. Then how should we prove that,
$$h_A^2+h_B^2 \geq 2h_C^2, \quad h_C^2+h_A^2\geq 2 h_B^2 \quad  and \quad h_B^2+h_C^2 \geq 2 h_A^2$$
I think that is true but only I can bounded like this:
$$BC^2 \geq h_B^2, \quad AB^2 \geq h_B^2 \quad \text{By Exterior Angle Theorem}$$
So, $BC^2+AB^2\geq h_B^2$ and I can't prove that $h_A^2+ h_C^2 \geq BC^2+AB^2$
 [/INST] We may use following inequality:
$$h_A+h_b+h_c< a+b+c=2p\space\space\space\space\space (1)$$
Also:
$$h_A+h_B+h_C> p$$
Let's suppose:
$$h_A+h_B+h_C\approx p+\frac p2=\frac{3p}2$$
Squaring both sides we get:
$$h_A^2+h_B^2=h_C^2+\frac 94 p^2-3p\cdot h_C-2h_Ah_B$$
So we must show that:
$$t=\frac 94 p^2-3p\cdot h_C-2h_Ah_B\geq h_C^2$$
If $h_A=h_B=\epsilon$ then $p\approx h_C$ and we have:
$$t=\frac94 \times 4h_C^2-3 h_C^2-\epsilon=6h_C^2-\epsilon$$
So there is possibility to have a triangle in which:
$h_A^2+h_B^2\geq h_C^2$
But there is no guarantee for this relation to be symmetric. for example Consider Right angle triangle with sides 3, 4, 5 where :
$h_B=4$, $h_C=3$ and $h_A=2.4$ we have:
$h_A^2+h_B^2=2.4^2+4^2=21.76>2\times 3^2=18$
$h_B^2+h_C^2=9+16=25> 2\times 2.4^2=11.54$
But:
$h_A^2+h_C^2=14.76< 2\times 4^2=32$
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