The perimeter of a triangle is

$\text{In\hspace{0.17em}}\Delta ABC$

$AD,BE,CF$are the altitudes.

$\therefore \text{In\hspace{0.17em}}ADB$is right angled triangle

 $AB>AD\to \left(1\right)(\because$ In any right angled triangle ,hypotenuse is the longest side)

Similarly)

$\text{In\hspace{0.17em}}BEC\text{\hspace{0.17em}\hspace{0.17em}is\hspace{0.17em}a\hspace{0.17em}right angled triangle},$

$BC>BE\to \left(2\right)$

In $ACF\mathrm{is}a\text{\hspace{0.17em}right angled\hspace{0.17em}triangle},$

$AC>CF\to \left(3\right)$

Adding (1),(2),(3), we get

$AB+BC+CA>AD+BE+CF$

$\Rightarrow \text{Perimeter\hspace{0.17em}of }\Delta ABC>\text{sum\hspace{0.17em}of\hspace{0.17em}altitudes}$