# An Inequality with Two Sets of Cevians

### Solution 1

Let be $x,y,z\in [0,1];$ $BA'=xa;$ $CA'=(1-x)a;$ $CB'=yb;$ $AB'=(1-y)b;$ $AC'=zc$; $BC'=(1-z)c.$

\displaystyle\begin{align} \frac{[A'B'C']}{[ABC]}&=\frac{[ABC]-[AB'C']-[BA'C']-[CA'B']}{[ABC]}\\ &=1-\frac{\frac{1}{2}z(1-y)bc\sin A }{\frac{1}{2}bc \sin A}-\frac{\frac{1}{2}x(1-z)ac\sin B}{\frac{1}{2}ac \sin B}\\ &\qquad\qquad-\frac{\frac{1}{2}y(1-x)ab\sin C}{\frac{1}{2}ab\sin C}\\ &=1-z(1-y)-x(1-z)-y(1-x)\\ &=1-(x+y+z)+xy+xz+yz \end{align}

To sum up:

(1)

$\displaystyle\frac{[A'B'C']}{[ABC]}=1-(x+y+z)+xy+xz+yz.$

From Ceva's theorem, $\displaystyle\frac{x}{1-x}\cdot\frac{y}{1-y}\cdot\frac{z}{1-z}=1.\,$ Further:

\begin{align} xyz&=(1-x)(1-y)(1-z)\\ &=(1-x-y+xy)(1-z)\\ &=1-z-x+xz-y+yz+xy-xyz \end{align}

So that

(2)

$2xyz=1-(x+y+z)+(xz+xy+yz).$

From (1),(2) it follows that

$\displaystyle\frac{[A'B'C']}{[ABC]}=2xyz=\frac{2(xa)(yb)(zc)}{abc},$

i.e.,

(3)

$\displaystyle\frac{[A'B'C']}{[ABC]}=\frac{2BA'\cdot CB'\cdot AC'}{abc}.$

Similarly,

(4)

$\displaystyle\frac{[A''B''C'']}{[ABC] }=\frac{2BA''\cdot CB''\cdot AC''}{abc}$

From (3) and (4) and the AM-GM inequality, it follows that

\displaystyle\begin{align} \frac{[A'B'C']}{[A''B''C'']}&=\frac{BA'\cdot CB'\cdot AC'}{BA''\cdot CB'' \cdot AC''}\\ &=\frac{BA'}{BA''}\cdot \frac{CB'}{CB''}\cdot \frac{AC'}{AC''}\\ &\leq \frac{1}{27}\left(\frac{BA'}{BA''}+\frac{CB'}{CB''}+\frac{AC'}{AC''}\right)^3. \end{align}

Finally,

$\displaystyle \frac{27[A'B'C']}{[ABC]}\leq \left(\frac{BA'}{BA''}+\frac{CB'}{CB''}+\frac{AC'}{AC''}\right)^3.$

### Solution 2

We will work in areal (normalized barycentric) coordinates. The coordinates of the triangle vertices are

$A\rightarrow(1,0,0),~B\rightarrow(0,1,0),~C\rightarrow(0,0,1).$

If the point of concurrency of $AA'$, $BB'$, and $CC'$ has coordinates $(p,q,r)$, then the coordinates of the end points of the cevians are

$\displaystyle A'\rightarrow\left(0,\frac{q}{q+r},\frac{r}{q+r}\right),~ B'\rightarrow\left(\frac{p}{p+r},0,\frac{r}{p+r}\right),~ C'\rightarrow\left(\frac{p}{p+q},\frac{q}{p+q},0\right).$

The lengths in terms of the sides of the triangle $\{a,b,c\}$ are,

$\displaystyle BA'=\frac{ar}{q+r},~ CB'=\frac{bp}{p+r},~ AC'=\frac{cq}{p+q},$

and

\displaystyle\begin{align} [A'B'C']&=&(\text{Determinant of the coordinates})\cdot[ABC] \\ &=&\frac{2pqr}{(p+q)(q+r)(r+p)}[ABC]. \end{align}

Let the point of concurrency of $AA''$, $BB''$, and $CC''$ be $(u,v,w)$. Thus, the inequality can be written as

$\displaystyle 27~\frac{pqr(u+v)(v+w)(w+u)}{uvw(p+q)(q+r)(r+p)}\leq\left[\frac{r(v+w)}{w(q+r)}+ \frac{p(w+u)}{u(r+p)}+\frac{q(u+v)}{v(p+q)}\right]^3,$

and follows from the AM-GM inequality.

### Acknowledgment

The problem, with a solution of his solution 1, was kindly communicated to me by Dan Sitaru. The problem has been published in the MathProblems Journal (Vol 6, n 2, p 592, #62.)

Note that the formula for the area of the cevian triangle has several incarnations each of which leads to a slightly different derivation of the penultimate step. Ultimately, all proofs reduce to an application of the AM-GM inequality. Solution 2 (by Amit Itagi) is one of the examples.