# The semiperimeter of an acute triangle is at least the perimeter of its orthic triangle

Let $ABC$ be an acute triangle. If $AD, BE,$ and $CF$ are the altitudes of the triangle $ABC$, prove that $$\text{perimeter of \triangle{DEF} \leq \text{semiperimeter of \triangle{ABC.}}}$$

We are to prove that the orthic triangle $DEF$'s perimeter is less than or equal to the semi perimeter of $\triangle{ABC}$. Of course, the perimeter of the orthic triangle is minimal among all inscribed triangles in $ABC$, but I don't think that helps us prove the inequality. Also I find it hard to relate the perimeter of the orthic triangle since we don't know very much about its sides. Okay, from your diagram, let $H_a=D,H_b=E$, and $H_c=F$. Notice cyclic quadrilateral $AFHE$. Center of a circle circumscribed around $AFHE$ is midpoint of $AH$ with radius of $\rho= \frac{AH}{2}$. It follows that $EF= 2 \rho \sin \alpha= AH \sin \alpha$ ($\alpha$ is angle at point $A$).

And $AH=\frac{AF}{\sin \beta}= \frac{b \cos \alpha } {\sin \beta}$ and since sine law says

$\frac{a } {\sin \alpha}= \frac{b} {\sin \beta}$

You get that

$EF= a \cos \alpha$

Similary you will find that $DF= b \cos \beta$ etc...

$DE+EF+FD = a \cos \alpha+ b \cos \beta +c \cos \gamma \leq \frac{1}{3} (a+b+c)(\cos \alpha + \cos \beta + \cos \gamma ) \leq \frac{a+b+c}{2}$

Since $\cos \alpha + \cos \beta + \cos \gamma \leq \frac{3}{2}$

If you know Nine-Point circle, it may be a little bit quick:

since $D,E,F$ are on the Nine-point circle, let the radius $R_0, R_1=2R_0,R_1$is the radius of $\triangle ABC$

let $m,n,p$ are the side length of $\triangle DEF,$ the the three angles are $\pi-2A,\pi-2B,\pi-2C$ $\dfrac{a+b+c}{sinA+sinB+sinC}=2R_1,\dfrac{m+n+p}{sin2A+sin2B+sin2C}=2R_0$

it remains $sin2A+sin2B+sin2C \le sinA+sinB+sinC$

you can use Chebyshev inequality get result as zezanjee's solution or

$sin2A+sin2B=2sinC cos(A-B)\implies sin2A+sin2B+sin2C=cos(B-C)sinA+cos(C-A)sinB+cos(A-B)sinC \le sinA+sinB+sinC$