Cleaner solutions exist, but... this problem isn't new!

This is not hard. Just take $X,Y$ the projections of $EF$ on $AB$, obvious, $EF\ge XY$, $XY=a-(p-a)(cosB+cosC)$. Now, take $a=y+z, b=x+z, c=x+y, x,y,z>0$. The inequality becomes $p\ge\displaystyle\sum\limits_{\text{cyc}}(p-a)(cosB+cosC)$, but $cosA=\frac{b^2+c^2-a^2}{2bc}$, so the inequality becomes ( in terms of $x,y,z$) $2(x+y+z)(x+y)(x+z)(y+z)\ge \displaystyle\sum\limits_{\text{cyc}}x[(x+z)(y+z)^2+(x+z)(x+y)^2-(x+z)^3-(x+y)^3+(y+z)^2(x+y)+(x+z)^2(x+y)]=\displaystyle\sum\limits_{\text{cyc}}x[(2x+y+z)(x^2+y^2+z^2+xy+xz+3yz)-(x+z)^3-(x+y)^3]=16xyz(x+y+z)$,and we have to prove $(x+y)(x+z)(y+z)\ge 8xyz$, well-known, so we are done.We have equality if and only if $EF=XY$ and the others are equal to their projections. In this case, it it easy to show that $\triangle{ABC}$ is equilateral. Just note that $EF=XY$ implies $EF \parallel BC$.

zephyredx wrote: ... It is easy to show that this projection also decreases the lengths ... Sorry, why ?! Thank you.

Virgil Nicula wrote: zephyredx wrote: ... It is easy to show that this projection also decreases the lengths ... Sorry, why ?! Thank you. There are probably a couple ways. I did it on the test by constructing a parallelogram (in other words, translating segment DE down onto the side) and doing some easy trig.

PP. Let $ABC$ be a triangle. Its incircle touches the sidelines $BC$ , $CA$ , $AB$ at the points $D$ , $E$ , $F$ respectively and its exincircles touch the sidelines $BC$ , $CA$ , $AB$ at the points $D'$ , $E'$ , $F'$ respectively. Prove that $\sum EF\ge \frac {2S}{R}$ and $\sum E'F'\ge s$ , where $2s=a+b+c$ . I"ll use in my proof some remarkable identities which are here with their proofs. $\blacktriangleright\ \sum a\cdot\cos A=$ $2R\cdot\sum \sin A\cos A=$ $R\cdot \sum\sin 2A=$ $4R\cdot\prod\sin A=$ $4R\cdot\frac {S}{2R^2}\implies$ $\boxed{\sum a\cdot\cos A=\frac {2S}{R}\ }\ \ (1)$ . $\blacktriangleright\ \sum a\cdot (\cos B+\cos C)=$ $\sum (b\cdot\cos C+c\cdot\cos B)=$ $\sum a\implies$ $\boxed{\ \sum a\cdot (\cos B+\cos C)=2s\ }\ \ (2)$ . $\blacktriangleright\ \sum\cos A=1-2\sin^2\frac A2+2\cos\frac {B+C}{2}\cos\frac {B-C}{2}=$ $1-2\sin\frac A2\left(\cos\frac {B+C}{2}-\cos\frac {B-C}{2}\right)=$ $1+4\prod\sin\frac A2=$ $1+4\sqrt{\prod \frac {(s-b)(s-c)}{bc}}=$ $1+\frac {4(s-a)(s-b)(s-c)}{abc}=$ $1+\frac {4sr^2}{4Rsr}\implies$ $\boxed{\ \sum\cos A=1+\frac rR\ }\ \ (3)$ . Otherwise, remark that $2s\cdot \sum\cos A=$ $\sum a\cdot \cos A+\sum a\cdot (\cos B+\cos C)\stackrel{(1)\wedge (2)}{=}$ $\frac {2S}{R}+2s\implies$ $\sum\cos A=1+\frac rR$ . Proof (Marius Bocanu). Denote the projection $\mathrm{pr}_{d}([XY])$ of $[XY]$ on the line $d$ . Therefore, $EF\ge \mathrm{pr}_{BC}([EF])=$ $\mathrm{pr}_{BC}([EA]+[AF])=$ $(s-a)\cdot (\cos B+\cos C)$ $\implies$ $\sum EF\ge \sum (s-a)\cdot (\cos B+\cos C)=$ $2s\sum\cos A-\sum a\cdot (\cos B+\cos C)\stackrel{(3)\wedge (2)}{=}$ $2s\left(1+\frac rR\right)-2s\implies$ $\boxed{\ EF+FD+DE\ge \frac {2S}{R}\ }$ . $\blacksquare\ E'F'\ge \mathrm{pr}_{BC}([E'F'])=a-\mathrm{pr}_{BC}(BF'+CE')=$ $a-(s-a)(\cos B+\cos C)\implies$ $\sum E'F'\ge \sum\left[a-(s-a)(\cos B+\cos C)\right]=$ $2s-2s\cdot\sum\cos A+\sum a\cdot (\cos B+\cos C)=$ $2s-2s\left(1+\frac rR\right)+2s=$ $2s\left(1-\frac rR\right)\ge 2s\left(1-\frac 12\right)$ $\implies$ $\boxed {\ E'F'+F'D'+D'E'\ge s\ }$ because $R\ge 2r$ .

zephyredx wrote:

Excuse me , do you mind explaining how to get the sum of the projected triangle? Is it $\triangle DXY$ in the attached figure? If so, I could just get that perimeter of $\triangle DXY < \triangle DEF$, but stuck in retrieving $\sum_{cyc} \frac{a^2}{b+c}$ being its perimeter, is there any trick?

Let the incircle touches $ AB , BC , CA $ at $ X,Y,Z $ respectively . Consider $ DE + XY \geq c $ , $ EF + YZ \geq a $ and $ FD + ZX \geq b $ It is enough to show that $ XY + YZ + ZX \leq \frac{a+b+c}{2} $ , but of course it isn't necessarily true , below is my attempt to show this fact . Suppose $ a \geq b \geq c $ , i find that $ AY \leq BZ \leq CX $ . Denote the mid-pts of $ XY , YZ , ZX $ by $ Z' , X' , Y' $ respectively . Then i find that $ IX' \geq IY' \geq IZ' $ . Using Chebyshev's inequality , $ XY + YZ + ZX = \frac{2}{r} \left( AY \cdot IX' + BZ \cdot IY' + CX \cdot IZ' \right) $ $ \leq \frac{2}{3r} ( AY + BZ + CX )( IX' + IY' + IZ' ) \leq \frac{2}{3r}\cdot \frac{a+b+c}{2} \cdot \frac{r+r+r}{2} = \frac{a+b+c}{2} $ ( The last part i have used Erdos'-Mordell inequality .)

Slightly less nice: Applying the Ravi substitution $a=x+y, b=y+z, c=z+x$, we have obviously $AE=x$, $AF=y$. Also $\cos{A}=\frac{z^2+y^2+xz-xy}{(x+z)(y+z)}=1-\frac{2xy}{(x+z)(y+z)}$ (from the Law of Cosines on triangle ABC) Thus from the law of cosines on triangle AEF, $EF=\sqrt{(x-y)^2+\frac{4x^2y^2}{(x+z)(y+z)}}$ We must then show that $\sum_{cyc} \sqrt{(x-y)^2+\frac{4x^2y^2}{(x+z)(y+z)}} \geq x+y+z$ To do this, it is sufficient to show that $\sqrt{(x-y)^2+\frac{4x^2y^2}{(x+z)(y+z)}} \geq \frac{3}{4}(x+y)-\frac{1}{2}z$ as summing that cyclically gives the desired result. Let $z=k(y+x)$ for some positive $k$. Note that if $k \geq \frac{3}{2}$, the RHS is negative so we're done. If $k \leq \frac{3}{2}$ then expanding shows that this is equivalent to: $16(x-y)^2(x+z)(y+z)+64x^2y^2 \geq (3x+3y-2z)^2(x+z)(y+z)$ NOW WE JUST TOTALLY EXPAND fortunately this is weaksauce and we get $-4z^4+8z^3(x+y)+z^2(19x^2-30xy+19y^2)+z(x+y)(7x^2-38xy+7y^2)+7xy(x+y)^2 \geq 0$ As $z=k(y+x)$, this is equivalent to $(-4k^4+8k^3)(y+x)^4+(y+x)^2[k^2(19x^2-30xy)+19y^2)+k(7x^2-38xy+7y^2)+7xy] \geq 0$ WLOG let $x+y=1$, $xy=m$ Then, we need to show that $(8k^3-4k^4)+k^2(19-8m)+k(7-52m)+7m \geq 0$ (completing the square for the $k^2$ and $k$ coefficients) or $m(68k^2+52k-7) \leq (-4k^4+8k^3+19k^2+7k)$. The $RHS$ is clearly non-negative, so if $68k^2+52k-7 \leq 0$ we are done. Else, we have $4xy \leq (x+y)^2 \implies m \leq \frac{1}{4}$ but $(1-2k)^2(7-4k-4k^2) \geq 0$ for $0 \leq k \leq \frac{3}{2}$ so $-16k^4 +32k^3+8k^2-24k+7 \geq 0 \implies -16k^4+32k^3+76k^2+28k \geq 68k^2+52k - 7$ $\implies \frac{-4k^4+8k^3+19k^2+7k}{68k^2+52k-7} \geq \frac{1}{4} \geq m$ as desired. Note that equality can only occur when $x=y$ and $k=\frac{1}{2}$, or $x=y=z$.

Let $M_a, M_b, M_c$ be the midpoints and $D', E', F'$ be the points of tangency of the incircle $(I)$ with sides $\overline{BC}, \overline{CA}, \overline{AB}$, respectively. Let $I_a, I_b, I_c$ be the excenters opposite $A, B, C$, and let $I_aD, I_bE, I_cF$ concur at the Bevan Point $V$ (the circumcenter of $\triangle I_aI_bI_c$). Let $R, r$ be the circumradius, inradius, respectively, of $\triangle ABC.$ We begin with a lemma: Lemma: $EF + E'F' \ge a.$ Proof of Lemma. Note that $IE' \perp AC, \; IF' \perp AB \implies A, I, E', F'$ are inscribed in a circle of diameter $\overline{AI}$, so by the Law of Sines, $\tfrac{E'F'}{\sin\angle E'AF'} = AI \implies E'F' = AI\sin A.$ Similarly, $VE \perp AC, \; VF \perp AB \implies A, V, E, F$ are inscribed in a circle of diameter $\overline{AV}$, so by the Law of Sines, $\tfrac{EF}{\sin\angle EAF} = AV \implies EF = AV\sin A.$ Now, here is the key: Note that $I$ is the orthocenter of $\triangle I_aI_bI_c$, so it is well-known that the reflection $I'$ of $I$ in $I_bI_c$ lies on the circumcircle of $\triangle I_aI_bI_c.$ By the triangle inequality, we have $VI' \le AV + AI' = AV + AI.$ But recall that the circumcircle of $\triangle ABC$ is just the nine-point circle of $\triangle I_aI_bI_c$, which means that the radius of $\odot (I_aI_bI_c)$ is just $2R$ (this follows from a homothety with center $I$ and ratio $2$). Then since $VI' = 2R$, it follows that \[EF + E'F' = (AI + AV)\sin A \ge 2R\sin A = a,\] where the last step follows from the Law of Sines. $\blacksquare$ Similarly, we find that \[FD + F'D' \ge b, \; DE + D'E' \ge c \implies (EF + FD + DE) + (E'F' + F'D' + D'E') \ge a + b + c.\] Therefore, it suffices to prove that $E'F' + F'D' + D'E' \le \tfrac{1}{2}(a + b + c).$ Recall that in any triangle $\triangle ABC$, the inequality $9R^2 \ge a^2 + b^2 + c^2$ holds. In fact, $a^2 + b^2 + c^2 \ge \tfrac{1}{3}\left(a + b + c\right)^2$ by AM-GM, so it follows that \[9r^2 \ge \tfrac{1}{3}\left(E'F' + F'D' + D'E'\right)^2 \implies E'F' + F'D' + D'E' \le 3\sqrt{3}r.\] (Note that we can also derive this inequality by using the fact that $E'F' = 2r\sin\angle E'D'F'$, and similar relations for the other sides, and then using Jensen's Inequality for the sine function) It remains to prove that $3\sqrt{3}r \le \tfrac{1}{2}(a + b + c) = s$, where $s$ denotes the semiperimeter of $\triangle ABC.$ Recall the following well-known facts: $[ABC] = rs = \sqrt{s(s - a)(s - b)(s - c)}.$ Hence, upon multiplying both sides by $s$, the desired inequality is equivalent to \[s^2 \ge 3\sqrt{3}rs = 3\sqrt{3}[ABC] = 9\sqrt{\frac{s}{3}(s - a)(s - b)(s - c)}.\] This follows from AM-GM: \[\frac{s}{3}(s - a)(s - b)(s - c) \le \left(\frac{\frac{s}{3} + (s - a) + (s - b) + (s - c)}{4}\right)^4 = \left(\frac{s}{3}\right)^4.\] $\square$

that the sequences $\{a, b, c\}$ and $\{\cos A, \cos B, \cos C\}$ are inversely sorted. Therefore, by Chebyhev's Inequality, \[a\cos A + b\cos B + c\cos C \le \frac{(a + b + c)(\cos A + \cos B + \cos C)}{3}.\]Thus, it remains to show that $\cos A + \cos B + \cos C \le 3 / 2.$ However, this follows immediately from Jensen's Inequality, or the identity $\cos A + \cos B + \cos C = 1 + r / R$ combined with Euler's Inequality: $2r \le R.$ $\square$

By the way Dukejukem, a messier but maybe faster way to prove the final inequality at the end is to use Schur's inequality. Anyways, consider the projection of $EF$ onto side $BC$. Suppose $E$ is projected to point $E'$ and $F$ is projected to point $F'$ on side $BC$. Basic triangle facts yield that $EF \ge E'F'=a-(s-a)(\cos B + \cos C)$. So we simply want \[ \sum_{cyc} a_(s-a)(\cos B + \cos C) \ge \frac{a+b+c}{2} \]. This rearranges to \[ a+b+c \ge 2a \cos A + 2b \cos B + 2c \cos C \]By LoS, it remains to show \[ \sin A + \sin B + \sin C \ge \sin 2A + \sin 2B + \sin 2C \]Observe that sum to product yields $\frac{\sin 2A + \sin 2B}{2} = \sin(A+B) \cos (A-B) \le \sin C$, from which summing yields the desired.

Let $E'F'$ be the projection of segment $EF$ onto $BC$. Clearly, $BF=CE=s-a$, so $EF\geq E'F'=a-BF'-CE'=a-(s-a)\cos B-(s-a)\cos C$. Summing over all of these inequalities, we have that $DE+EF+FD=a+b+c-\sum(s-a)(\cos B+\cos C)=a+b+c-\sum\cos A(s-b+s-c)$. Since $s-b+s-c=a$, we have that the perimeter of $DEF$ is at least $a+b+c-a\cos A-b\cos B-c\cos C$. We want to show that this is at least $\frac{a+b+c}{2}$, which is equivalent to $\frac{a+b+c}{2}\geq a\cos A+b\cos B+c\cos C$. By the Law of Cosines, $\cos A=\frac{b^2+c^2-a^2}{2bc}$, so substituting it remains to show that $abc(a+b+c)\geq\sum a^2(b^2+c^2-a^2)$, or $abc(a+b+c)\geq 2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4=16[ABC]^2$ by Heron's formula. Making the Ravi Substitution, it remains to show that $2(x+y)(y+z)(z+x)(x+y+z)\geq16xyz(x+y+z)$, and this immediately follows from the classic inequality $(x+y)(y+z)(z+x)\geq8xyz$.

MellowMelon wrote: Let $ABC$ be a triangle. Its excircles touch sides $BC, CA, AB$ at $D, E, F$, respectively. Prove that the perimeter of triangle $ABC$ is at most twice that of triangle $DEF$. Let $D'E'F'$ be the contact triangle, $I, J, O$ be the incenter, circumcenter and Bevan point, respectively. Note that $\triangle D'E'F'$ is the pedal triangle of $J$ in $\triangle ABC$. Claim: $EF+E'F' \geqslant BC$. (Proof) Note that $O$ is the midpoint of $IJ$ so $AI+AJ \geqslant 2AO$. Multiplying by $\sin A$ on both sides gives the claim. $\square$ Now, it suffices to show $$\sum_{\text{cyc}} E'F' \leqslant \frac{1}{2} \sum_{\text{cyc}} BC \iff \sum_{\text{cyc}} (b+c-a)\sin \frac{A}{2} \leqslant \sum_{\text{cyc}} \frac{a}{2}.$$By Tchebychef's inequality, we have $$\sum_{\text{cyc}} a \left(\frac{1}{2}+2\sin \frac{A}{2}\right) \geqslant \frac{1}{3} \left(\sum_{\text{cyc}} a \right) \left(\sum_{\text{cyc}} \left(\frac{1}{2}+2\sin \frac{A}{2}\right) \right) \geqslant \left(\sum_{\text{cyc}} a \right) \left(\sum_{\text{cyc}} \sin \frac{A}{2} \right).$$Here we used the well-known fact that $$\sum_{\text{cyc}} \sin \frac{A}{2} \leqslant \frac{3}{2},$$for angles of a triangle. It is clear to see that the last inequality rearranges to the desired one. $\blacksquare$

Let the incircle touch the sides at $D'$, $E'$, and $F'$. For any triangle $XYZ$, let $p(XYZ)$ denote its perimeter. Claim: We have \[p(D'E'F')\le\frac{p(ABC)}{2}.\] Proof: Note that $E'F'=r\cos(A/2)$, so by Jensen's inequality, we have \[\sum_{\mathrm{cyc}}E'F'\le r\cdot 3\sqrt{3}.\]Note that $rs=\sqrt{s(s-a)(s-b)(s-c)}$, so \[r=\frac{1}{\sqrt{s}}\sqrt{(s-a)(s-b)(s-c)}\le \frac{1}{\sqrt{s}}\sqrt{(s/3)^3}\]by the AM-GM inequality. Thus, \[p(D'E'F')\le s,\]as desired. $\blacksquare$ We claim that this solves the problem. In particular, we have the following lemma. Lemma: We have \[EF+E'F'\ge BC.\] Proof: Note that since $E$ and $E'$ are reflections over the midpoint of $AC$, and $F$ and $F'$ are reflections over the midpoint of $AB$, we have that $|\pi(EF)|+|\pi(E'F')|=BC$. Here $\pi$ is the projection map onto $BC$. But $EF\ge|\pi(EF)|$ and $E'F'\ge|\pi(E'F')|$, so the lemma is proved. $\blacksquare$ Combining the lemma and the claim immediately gives the desired result.

Consider the projections $E_A, F_A$ of points $E$ and $F$ onto side $BC$. Clearly, we have $EF\ge E_AF_A$. Also, \[E_AF_A=a-(s-a)(\cos B+\cos C).\]So, summing cyclically, we get that the perimeter of triangle $DEF$ is at least \[(a+b+c)-(a\cos A+b\cos B+c\cos C).\]It now suffices to show that \[a+b+c\ge 2(a\cos A+b\cos B+c\cos C).\]In fact, I claim that we have $c\ge a\cos A+b\cos B$, which yields the desired result. We wish to show that \[c\ge \frac{a(b^2+c^2-a^2)}{2bc}+\frac{b(a^2+c^2-b^2)}{2ac},\]which is equivalent to \[2abc^2\ge 2a^2b^2+a^2c^2+b^2c^2-a^4-b^4.\]Rearranging, we get \[(a^2-b^2)^2\ge c^2(a-b)^2,\]which is true because $c\le a+b$.

We prove the stronger condition: For any triangle $XYZ$ with $X,Y,Z$ on $BC,CA,AB$ respectively, the perimeter of $XYZ$ is at least half of $ABC$. Lemma: For fixed points $P,Q$ and a fixed line $l$, let $R$ be a point varying along $l$. Then, the perimeter of triangle $PQR$ is minimal if and only if the acute angles that $PR$ and $QR$ form with $l$ are equal. Proof: Reflect $Q$ over $l$ to be $Q'$. Since $PQ$ is fixed, we seek to minimise $PR+QR=PR+Q'R$, if and only if $P,Q',R$ are collinear. Vertically opposite angles are equal, reflecting $Q'$ back to $Q$ proves the lemma. To prove the stronger condition, we first claim that the only triangle $XYZ$ that satisfies equality is the medial triangle. Its perimeter is indeed half of $ABC$, so we now seek to prove that the medial triangle has minimal perimeter. Suppose there exist some triangle $TUV$ with $T,U,V$ on $BC,CA,AC$ that has minimal perimeter (i.e. $P_{TUV}\leq P_{XYZ}$ where $XYZ$ is the medial triangle). Let $\alpha=\angle A$, and define $\beta$ and $\gamma$ similarly. Set $\angle BTV=\angle CTU=x$, $\angle CUT=\angle AUV=y$ and $\angle AVU=\angle AVT=z$, as if one of the equalities do not hold, , we can minimise the perimeter further using our lemma, a contradiction. Angle chasing now gives $$2x+2y+2z+\alpha+\beta+\gamma=540^{\circ}$$so $$x+y+z=180^{\circ}$$Angle sum also gives $$\alpha+y+z=180^{\circ}$$Hence, $$x=\alpha, y=\beta, z=\gamma$$This gives $\Delta XYZ\sim\Delta TUV$ and their corresponding sides are parallel. It is easy to see that this is not possible unless $T,U,V$ are indeed midpoints, so $\Delta TUV$ is the medial triangle. Hence, $P_{DEF}\geq P_{XYZ}=\frac{1}{2}P_{ABC}$, completing the proof.

Good solution

MariusBocanu wrote: This is not hard. Just take $X,Y$ the projections of $EF$ on $AB$, obvious, $EF\ge XY$, $XY=a-(p-a)(cosB+cosC)$. Now, take $a=y+z, b=x+z, c=x+y, x,y,z>0$. The inequality becomes $p\ge\displaystyle\sum\limits_{\text{cyc}}(p-a)(cosB+cosC)$, but $cosA=\frac{b^2+c^2-a^2}{2bc}$, so the inequality becomes ( in terms of $x,y,z$) $2(x+y+z)(x+y)(x+z)(y+z)\ge \displaystyle\sum\limits_{\text{cyc}}x[(x+z)(y+z)^2+(x+z)(x+y)^2-(x+z)^3-(x+y)^3+(y+z)^2(x+y)+(x+z)^2(x+y)]=\displaystyle\sum\limits_{\text{cyc}}x[(2x+y+z)(x^2+y^2+z^2+xy+xz+3yz)-(x+z)^3-(x+y)^3]=16xyz(x+y+z)$,and we have to prove $(x+y)(x+z)(y+z)\ge 8xyz$, well-known, so we are done.We have equality if and only if $EF=XY$ and the others are equal to their projections. In this case, it it easy to show that $\triangle{ABC}$ is equilateral. Just note that $EF=XY$ implies $EF \parallel BC$. What motivates the projection?

Bump. And user @NoctNight, your last step is not true. Consider orthic triangle

By Ravi Substitution, we have $AE=AF=x$, $BF=BD=y$, $DC=CE=z$. Then, by Law of Cosines on $\triangle ABC$, we have that \[\cos C=\frac{(x+z)^2+(y+z)^2-(x+y)^2}{2(x+z)(y+z)}\]\[ =\frac{xz+z^2+yz-xy}{(x+z)(y+z)} \] Then, by Law of Cosines on $\triangle CDE$, we have that \[ ED=z\sqrt{2(1-\frac{xz+z^2+yz-xy}{(x+z)(y+z)}}. \]and similarly for the other sides. Thus, equivalantly, we can show that \[ \sum_{\text{cyc}} z\sqrt{2(1-\frac{xz+z^2+yz-xy}{(x+z)(y+z)})} \leq x+y+z. \]Or, equivalantly (by rearrangement) that \[ \sum_{\text{cyc}} \sqrt{2(1-\frac{xz+z^2+yz-xy}{(x+z)(y+z)})} \leq 3 \]Evaluating the inner part of the square root, we wish to show that \[ \sum_{\text{cyc}} 2 \sqrt{\frac{xy}{(x+z)(y+z)}} \leq 3 \]Dividing by $2$ on both sides, we wish to show that \[ \sum_{\text{cyc}} \sqrt{\frac{xy}{(x+z)(y+z)}} \leq \frac{3}{2} \]which follows from AM-GM since each term of the LHS is $\leq \sqrt{\frac{1}{4}}$, or the whole thing is $\leq \frac{3}{2}$.

Let $D'E'F'$ be the intouch triangle of $ABC$. By projecting onto the sides, it follows that $p(D'E'F') + p(DEF) \geq p(ABC)$. Then it suffices to show that $p(D'E'F') \leq p(ABC)$. This is doable by direct computation: $$\sum E'F' = \sum (b+c-a) \sin \frac A2 \leq \frac 13 (a+b+c)\left(\sin \frac A2+\sin \frac B2 + \sin \frac C2\right) \leq \frac 12 (a+b+c)$$by Jensen.