Problem 4020 from Crux Mathematicorum
Statement
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.
Solution 4
Lemma
Let $AX,\;$ $BY,\;$ $CZ\;$ be concurrent cevians such that $\displaystyle\frac{XB}{XC}=\frac{z}{y},\;$ $\displaystyle\frac{YC}{YA}=\frac{x}{z},\;$ and $\displaystyle\frac{ZA}{ZB}=\frac{y}{x},\;$ where $x,y,z\gt 0.\;$ Then
$\displaystyle [XYZ]=\frac{2xyz}{(x+y)(y+z)(z+x)}\cdot S,$
where $S=[\Delta ABC].$
Indeed, we find that $\displaystyle\frac{XB}{BC}=\frac{z}{y+z},\;$ implying $\displaystyle XB=\frac{az}{y+z}.\;$ Similarly, $\displaystyle ZB=\frac{cx}{x+y}.\;$ Hence,
$\displaystyle\begin{align} 2[\Delta BXZ]&=\frac{az}{y+z}\cdot\frac{cx}{x+y}\cdot\sin B\\ &=\frac{z}{y+z}\cdot\frac{x}{x+y}\cdot ac\sin B\\ &=\frac{2xz}{(x+y)(y+z)}\cdot S. \end{align}$
It follows that $\displaystyle [\Delta BXZ]=\frac{xz}{(x+y)(y+z)}\cdot S.\;$ Similarly, $\displaystyle [\Delta AZY]=\frac{yz}{(x+y)(z+x)}\cdot S\;$ and $\displaystyle [\Delta CYX]=\frac{xy}{(y+z)(z+x)}\cdot S.\;$ Now
$\displaystyle\begin{align}[\Delta XYZ]&=S-\left([\Delta BXZ]+[\Delta AZY] + [\Delta CYX]\right)\\ &=\frac{2xyz}{(x+y)(y+z)(z+x)}\cdot S. \end{align}$

Back to the main proof, we apply the lemma for $X=M,\;$ $Y=N,\;$ $Z=P\;$ and $\displaystyle x=\frac{1}{s-a},\;$ $\displaystyle y=\frac{1}{s-b},\;$ $\displaystyle z=\frac{1}{s-c}\;$ to obtain
$\displaystyle [\Delta MNP]=\frac{2(s-a)(s-b)(s-c)}{abc}\cdot S.$
Similarly, for $X=D,\;$ $Y=E,\;$ $Z=F\;$ and $x=a,\;$ $y=b,\;$ $z=c\;$ we obtain
$\displaystyle [\Delta DEF]=\frac{2abc}{(a+b)(b+c)(c+a)}\cdot S.$
We need to prove that
$\displaystyle \frac{(s-a)(s-b)(s-c)}{abc}\le \frac{abc}{(a+b)(b+c)(c+a)}.$
By Heron's formula $\displaystyle (s-a)(s-b)(s-c)=\frac{S^2}{s};\;$ also, $abc=4RS\;$ and $\displaystyle s=4R\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}.$
But
$\displaystyle a+b=2R(\sin A+\sin B)=4R\cos\frac{C}{2}\cos\frac{A-B}{2},\\ \displaystyle b+c=2R(\sin B+\sin C)=4R\cos\frac{A}{2}\cos\frac{B-C}{2},\\ \displaystyle c+a=2R(\sin C+\sin A)=4R\cos\frac{B}{2}\cos\frac{C-A}{2}.$
Hence our inequality is equivalent to
$\displaystyle\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}\cos\frac{A-B}{2}\cos\frac{B-C}{2}\cos\frac{C-A}{2}\le\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$
which is obviously true.
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. Solution 4 is by Leonard Giugiuc and Daniel Sitaru.
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- An inequality with Three Points $\displaystyle\left(\sum_{P\in\{O,I,G\}}\sum_{cycl}\left(\frac{[\Delta APB]}{[\Delta ABC]}+\frac{[\Delta ABC]}{[\Delta APB]}\right)^2\ge 100\right)$
- An Inequality with One Tangent and Six Sines $\left(\displaystyle \frac{\tan A}{\sin B+5\sin C}+\frac{\tan B}{\sin C+5\sin A}+\frac{\tan C}{\sin A+5\sin B}\gt\frac{1}{2}\right)$
- An Inequality with Tangents and Sides $\left(\displaystyle \frac{a^2}{\tan B+\tan C}+\frac{b^2}{\tan C+\tan A}+\frac{c^2}{\tan A+\tan B}\leq sR\right)$
- An Inequality with Sides, Altitudes, Angle Bisectors and Medians $\left(\displaystyle \left(\frac{h_b}{m_a}+\frac{h_c}{m_b}+\frac{h_a}{m_c}\right)\left(\frac{h_b}{\ell_a}+\frac{h_c}{\ell_b}+\frac{h_a}{\ell_c}\right)\ge\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)^2\right)$
- An Inequality of the Areas of Triangles Formed by Circumcenter And Orthocenter $\left(\displaystyle \sum_{cycl}\sqrt[n]{[\Delta OAB]}\ge\sum_{cycl}\sqrt[n]{[\Delta HAB]}\right)$
- An Inequality with Angle Trisectors $\left(\displaystyle \frac{AE}{AB}+\frac{AF}{AC}\lt 2\right)$
- An Inequality for Sides and Area $\left(\displaystyle \sum_{cycl}\frac{(a^2-ab+b^2)^2}{a^2+4ab+b^2}\ge\frac{2S}{\sqrt{3}}\right)$
- A Cyclic Inequality in Triangle for Integer Powers $\left(\displaystyle \sum_{cycl}\frac{a^{n+1}}{b+c-a}\ge\sum_{cycl}a^n\right)$
- R and r When G on Incircle $(8R\ge 25r)$
- An Inequality for Cevians And The Ratio of Circumradius and Inradius $\left(\displaystyle \frac{XB}{XY}\cdot\frac{YC}{YZ}\cdot\frac{ZA}{ZX}\le\frac{R}{2r}\right)$
- Centroid on the Incircle in Right Triangle $\left(648Rr\ge 25 (a^2+b^2+c^2)\right)$
- An Inequality with Cotangents And the Circumradius $\left(\displaystyle \sum_{cycl}a^2\cot B\cot C\le 4R^2\right)$
- An Inequality with Inradius and Excenters $\left(\displaystyle \sum_{cycl}\frac{1}{II_a^2}+\sum_{cycl}\frac{1}{I_aI_b^2}\le\frac{1}{4r^2}\right)$
- An Inequality with Inradius and Side Lengths $\left(\displaystyle \sum_{cycl}(b+c-a)^2\cdot\sum_{cycl}(b+c-a)^3\ge 2592\sqrt{3}r^5\right)$
- An Inequality with Exradii and an Altitude $\left(\displaystyle \sqrt{\frac{1}{r_b^2}+\frac{1}{r_c}+1}+\sqrt{\frac{1}{r_c^2}+\frac{1}{r_b}+1}\ge 2\sqrt{\frac{1}{h_a^2}+\frac{1}{h_a}+1}\right)$
- Leuenberger's Inequality for Medians, Inradius and Circumradius $\left(\displaystyle m_a+m_b+m_c\le 4R+r\right)$
- Adil Abdulayev's Inequality With Angles, Medians, Inradius and Circumradius $\left(\displaystyle \frac{A}{m_a}+\frac{B}{m_b}+\frac{C}{m_c}\le \frac{3\pi}{4R+r}\right)$
- An Inequality with Sides, Cosines, and Semiperimeter $\left(\displaystyle \sum_{cycl}a^2(b\cos B+c\cos C)\le \frac{8s^3}{9}\right)$
- Seyran Ibrahimov's Inequality $\left(\displaystyle \sqrt{3}s\cdot\sum_{cycl}m_a\le 20R^2+r^2\right)$
- An Inequality in Triangle, with Sides and Medians III $\left(\displaystyle \sum_{cycl}\frac{(m_b+m_c-m_a)^3}{m_a}\ge\frac{3}{4}(a^2+b^2+c^2)\right)$
- Leo Giugiuc's Inequality in Triangle, Solely with Cotangents $\left(\cot A +\cot B+\cot C\ge 2\sqrt{2}-1\right)$
- An Inequality in Triangle with Side Lengths and Circumradius $\left(\displaystyle\displaystyle\frac{ab}{\sqrt{a^2+b^2}}+\frac{bc}{\sqrt{b^2+c^2}}+\frac{ca}{\sqrt{c^2+a^2}}\le\frac{3\sqrt{6}R}{2}\right)$
- All Trigonometric Inequality in Triangle $\left(\displaystyle 3\sum_{cycl}\cos A\ge 2\sum_{cycl}\sin A\sin B\right)$
- An Inequality with Two Sets of Cevians $\left(\displaystyle \frac{27[A'B'C']}{[A''B''C'']}\leq \Bigr(\frac{BA'}{BA''}+\frac{CB'}{CB''}+\frac{AC'}{AC''}\Bigr)^3\right)$
- An Inequality with the Most Important Cevians $\left(\displaystyle \frac{m_am_bm_c}{h_ah_bh_c}\ge\frac{\ell_a^2+\ell_b^2+\ell_c^2}{\ell_a\ell_b+\ell_b\ell_c+\ell_c\ell_a}\right)$
- An Inequality in Triangle with a Constraint $(a\sqrt{2}\ge b+c)$
- Problem 11984 From the American Mathematical Monthly $\left(\displaystyle a^6+b^6+c^6\ge 5184\cdot r^6\right)$
- A Long Cyclic Inequality of Degree 4 $\left(\displaystyle 4\cdot\sum_{cycl}ab\cdot\sum_{cycl}a-\left(\sum_{cycl}a\right)^3\ge\frac{\displaystyle 3\sum_{cycl}ab\left[4\sum_{cycl}ab-\left(\sum_{cycl}a\right)^2\right]}{\displaystyle \sum_{cycl}a}\right)$
- An Inequality of Degree 3 with Inradius $\left(\displaystyle \sum_{cycl}(a+b-c)^3+24abc\ge 648\sqrt{3}r^3\right)$
- A Cyclic Inequality from the 6th IMO, 1964 $\left(\displaystyle \sum_{cycl}a^2(b+c-a)\le 3abc\right)$
- An Area Inequality in Right Triangle $\left(\displaystyle \frac{[\Delta ABD]+[\Delta ACE]}{[\Delta ADE]}\ge\sqrt{2}\right)$
- An Angle Inequality in Triangle with Perpendicular Medians $\left(\displaystyle \cos A\ge \frac{4}{5}\right)$
- An Inequality in Triangle form the 1996 APMO $\left(\sqrt{a+b-c}+\sqrt{b+c-a}+\sqrt{c+a-b}\le\sqrt{a}+\sqrt{b}+\sqrt{c}\right)$
- An Inequality in a Nonobtuse Triangle $\left(R\sqrt{2}\le h_a\right)$
- An Inequality in Triangle with Radicals, Semiperimeter and Inradius $\left(\displaystyle \sqrt{\frac{a+b}{s-b}}+ \sqrt{\frac{b+c}{s-c}} +\sqrt{\frac{c+a}{s-a}}\le \frac{\sqrt{a^2+b^2+c^2}}{r}\right)$
- Dan Sitaru's Inequality with Radicals and Cosines $\left(\displaystyle (a^2+b^2+c^2)^3\ge 6^3(abc)^2\cos A\cos B\cos C\right)$
- Lorian Saceanu's Cyclic Inequalities with Three Variables $\left(\displaystyle 2+\sum_{cycl}\frac{a}{b}\ge \sum_{cycl}\frac{a}{c}+2\frac{ab+bc+ca}{a^2+b^2+c^2}\right)$
- An Inequality for the Tangent to the Incircle $\left(\displaystyle DE\le\frac{1}{8}(AB+BC+CA)\right)$
- Lorian Saceanu's Sides And Angles Inequality $\left(\displaystyle \frac{\pi}{3}\le \frac{a\alpha + b\beta + c\gamma}{a+b+c}\le \arccos\left(\frac{r}{R}\right)\right)$
- An Inequality in Triangle with Radicals, Semiperimeter, Incenter and Inradius $\left(\displaystyle \frac{AI+BI+CI}{r}+3\ge\left(\sum_{cycl}\sqrt{s-a}\right)\left(\sum_{cycl}\frac{1}{\sqrt{s-a}}\right)\right)$
- An Inequality in Triangle for Side Lengths, Cycled in Two Ways $\left(\displaystyle 3\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-1\right)\ge 2\left(\frac{b}{a}+\frac{a}{c}+\frac{c}{b}\right)\right)$
- Lorian Saceanu's Inequality for All Triangles $\left(\displaystyle \sin 2A+\sin 2B+\sin 2C\ge 4\sin 2A\cdot\sin 2B\cdot\sin 2C\right)$

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