Problem 4020 from Crux Mathematicorum

Statement

Problem 4020 from Crux Mathematicorum

Solution 1

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.

 

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