# Problem 4020 from Crux Mathematicorum

Denote by $\alpha,\;$ $\beta,\;$ $\gamma\;$ the angles $BAC,\;$ $ABC,\;$ and $ACB,\;$ respectively; and let $r\;$ be the inradius of $\Delta ABC. From the quadrilateral$PIMB,\;\angle PIM = 180^{\circ}-\beta,\;$whence$\displaystyle [PIM]=\frac{PI\cdot MI}{2}\sin\angle PIM=\frac{r^2}{2}\sin (180^{\circ}-\beta)=\frac{r^2}{2}\sin \beta.$Similarly, we calculate$[MIN]\;$and$[NIP],\;$and get$\displaystyle\begin{align} [MNP]&=[PIM]+[MIN]+[NIP]\\ &=\frac{r^2}{2}\left(\sin\alpha +\sin\beta +\sin\gamma\right). \end{align}\;\;\;\;\;\text{(1)}$On the other hand, we have$\displaystyle\angle FID=\angle AIC=180^{\circ}-\frac{\alpha}{2}-\frac{\gamma}{2},\;$and so$\displaystyle [FID]=\frac{ID\cdot IF}{2}\sin\angle FID=\frac{ID\cdot IF}{2}\sin\angle\frac{\alpha+\gamma}{2}.$Similarly, calculate$[\Delta EIF]\;$and$[\Delta DIE].\;$We have$\displaystyle\begin{align} [DEF]&=[DIE]+[EIF]+[FID]\\ &=\frac{ID\cdot IE}{2}\sin\frac{\alpha+\beta}{2}+\frac{IE\cdot IF}{2}\sin\frac{\beta+\gamma}{2}+\frac{ID\cdot IF}{2}\sin\frac{\alpha+\gamma}{2}. \end{align}\;\;\;\;\;\text{(2)}$Triangles$PIF,\;MID,\;$and$NIF\;$are all right-angled, implying that$IF\ge r,\;ID\ge r,\;IE\ge r.\;$Hence, from (2),$\displaystyle [\Delta DEF]\ge \frac{r^2}{2}\left(\sin\frac{\alpha+\beta}{2}+\sin\frac{\beta+\gamma}{2}+\sin\frac{\alpha+\gamma}{2}\right).$Comparing this to (1), in order to have$[MNP]\le [DEF],\;$suffice it to show that$\displaystyle \sin\alpha +\sin\beta +\sin\gamma\le\sin\frac{\alpha+\beta}{2}+\sin\frac{\beta+\gamma}{2}+\sin\frac{\alpha+\gamma}{2}.\;\;\;\;\;(3)$However, we know that$\displaystyle \sin\alpha +\sin\beta=2\sin\frac{\alpha +\beta}{2}\cos\frac{\alpha-\beta}{2}\le 2\sin\frac{\alpha +\beta}{2}.$Similarly,$\displaystyle \sin\beta +\sin\gamma\le 2\sin\frac{\beta +\gamma}{2}\;$and$\displaystyle \sin\alpha +\sin\gamma\le 2\sin\frac{\alpha +\gamma}{2}.\;$Adding up proves (3) and, therefore, the required inequality. The equality occurs for$\alpha=\beta=\gamma.$### Solution 2 With$R\;$the circumradius,$O\;$the circumcenter,$s\;$the semiperimeter, we have the following sequence:$\displaystyle [\Delta MNP]=\frac{R^2-OI^2}{4R^2}\cdot [ABC]=\frac{2Rr}{4R^2}=\frac{r}{4R^2}\cdot [\Delta ABC]\\ \displaystyle [\Delta DEF]=\frac{2abc}{(a+b)(b+c)(c+a)}\cdot [ABC]\\ \displaystyle \frac{r}{2R}\le \frac{2abc}{(a+b)(b+c)(c+a)}\\ \displaystyle\begin{align} \frac{r}{2R}&=\frac{[\Delta ABC]}{s}:\frac{2abc}{4[\Delta ABC]}=\frac{2[\Delta ABC]^2}{sabc}\\ &=\frac{2s(s-a)(s-b)(s-c)}{sabc}\\ &=\frac{2(s-a)(s-b)(s-c)}{abc}. \end{align}$Thus suffice it to prove that$\displaystyle\frac{2(s-a)(s-b)(s-c)}{abc}\le\frac{2abc}{(a+b)(b+c)(c+a)},$which is equivalent to$(s-a)(s-b)(s-c)(a+b)(b+c)(c+a)\le (abc)^2.$Let$s-a=x\gt 0,\;s-b=y\gt 0,\;s-c=z\gt 0,\;$then$a=y+z,\;b=x+z,\;c=x+y,\;$and the above inequality becomes$xyz(x+y+2z)(x+2y+z)(2x+y+z)\le (x+y)^2(y+z)^2(z+x)^2.$By the AM-GM inequality,$\displaystyle\begin{align}(x+y+2z)(x+2y+z)&(2x+y+z)\\ &\le \left(\frac{(x+y+2z)+(x+2y+z)+(2x+y+z)}{3}\,\right)^3\\ &=\left(\frac{4(x+y+z)}{3}\right)^3\\ \end{align}$Thus, if we can prove that$\displaystyle xyz\left(\frac{4(x+y+z)}{3}\right)^3\le (x+y)^2(y+z)^2(z+x)^2$the problem will be solved. The inequality is equivalent to$\displaystyle \ln x+\ln y+\ln z+3\ln\frac{x+y+z}{3}\le 2\ln\frac{x+y}{2}+2\ln\frac{y+z}{2}+2\ln\frac{z+x}{2}.$This is Tiberiu Popoviciu's inequality$\displaystyle f(x)+f(y)+f(z)+3f\left(\frac{x+y+z}{3}\right)\ge 2f\left(\frac{x+y}{2}\right)+2f\left(\frac{y+z}{2}\right)+2f\left(\frac{z+x}{2}\right)$for the convex function$f(x)=-\ln (x)\;$which proves the result. ### Solution 3 We'll use the barycentric coordinates. The area of a triangle$T\;$whose vertices have homogeneous barycentric coordinates$(x_1:y_1:z_1),\;(x_2:y_2:z_2),\;(x_3:y_3:z_3),\;$(relative to$\Delta ABC)\;$is given by$\displaystyle [T]=\frac{[\Delta ABC]}{(x_1+y_1+z_1)(x_2+y_2+z_2)(x_3+y_3+z_3)}\left|\begin{array} \,x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\end{array}\right|.$For a triangle formed by the feet of three cevians that concur at point$(x:y:z)\;$that formula becomes$\displaystyle [T]=\frac{[\Delta ABC]}{(y+z)(z+x)(x+y)}\left|\begin{array} \,0&y&z\\x&0&z\\x&y&0\end{array}\right|=\frac{2xyz[\Delta ABC]}{(y+z)(z+x)(x+y)}.$In particular, since$I=(a:b:c),\;\displaystyle [\Delta DEF]=\frac{2abc[\Delta ABC]}{(a+b)(b+c)(c+a)}.$The vertices of the intouch triangle$MNP\;$have barycentric coordinates$(0:s-b:s-c),\;(s-a:0:s-c),\;(s-a:s-b:0),\;$the corresponding cevians meet at the Gergonne point, and, therefore,$\displaystyle [\Delta MNP]=\frac{2(s-a)(s-b)(s-c)[\Delta ABC]}{abc}$which reduces the problem to proving$\displaystyle \frac{(s-a)(s-b)(s-c)}{abc}\le\frac{abc}{(a+b)(b+c)(c+a)}.$This, in turn, is equivalent to$\displaystyle \frac{(a+b)^2(s-a)(s-b)}{abc^2}\cdot\frac{(b+c)^2(s-b)(s-c)}{a^2bc}\cdot\frac{(c+a)^2(s-c)(s-a)}{abc^2}\le 1.$However, as we know, say,$\displaystyle \frac{(b+c)^2(s-b)(s-c)}{a^2bc}=\cos^2\frac{\beta -\gamma}{2},$which reduces the required inequality to the trivial$\displaystyle \cos^2\frac{\alpha -\beta}{2}\cdot\cos^2\frac{\beta -\gamma}{2}\cdot\cos^2\frac{\gamma -\alpha}{2}\le 1,\$

with equality only when all three angles are equal.

### Acknowledgment

The problem has been proposed by Leonard Giugiuc and Daniel Sitaru. The solution was published in Crux Mathematicorum, VOLUME 42, NO. 2 February / Février 2016, with the following editorial remark:

We received eleven submissions, of which nine were correct and complete. We present the solution by Sefket Arslanagic, slightly modified by the editor.

Thus, Solution 1 is by Sefket Arslanagic. I am grateful to Marian Dinca who communicated to me the problem, Sefket Arslanagic's solution, and a solution (Solution 2) of his own. Solution 3 is by yours truly.