Given concurrent cevians $AD$, $BE$, $CF$ of $\triangle ABC$, show $\text{area }\triangle DEF\leq \tfrac14\text{}\triangle ABC$

Question

Let $M$ be a point in $\triangle ABC$. Lines $AM$, $BM$, $CM$ respectively intersect $BC$, $CA$, $AB$ at $D$, $E$, $F$. Prove that $$S_{DEF}\leq \tfrac{1}{4}S_{ABC}$$ where $S_{XYZ}$ indicates the area of $\triangle XYZ$.

Answer 1

$S_{DEF}\leq \frac{1}{4}S_{ABC}$ can be rewritten as $4S_{DEF}\leq S_{ABC}$.

You can scale this triangle so that its circumradius is equal to $\frac{1}{4}$, making $S_{ABC} = abc = (a_1+a_2)(b_1+b_2)(c_1+c_2)$. This is because $A = \frac{abc}{4R}$.

Start by finding the areas of $\triangle EDC$, $\triangle AFE$, and $\triangle FBD$.

$S_{AFE} = \frac{1}{2}c_1b_2\sin A$. Since $2R = \frac{a}{\sin A}$, $\frac{1}{2} = \frac{a}{\sin A}$, and $\sin A=2a$. Plugging this into the area you get, $S_{AFE} = c_1b_2(a_1+a_2)$

With this method, $S_{CDE} = a_2b_1(c_1+c_2)$ and $S_{FBD} = a_1c_2(b_1+b_2)$.

Whe combined, the expressions for these areas contain all the terms from the expression of $S_{ABC}$ except $(a_1b_1c_1)$ and $(a_2b_2c_2)$. From this, you get the expression $S_{DEF}= S_{ABC}-S_{CDE}-S_{FBD}-S_{AFE} = (a_2b_2c_2) + (a_1b_1c_1)$

Ceva's Theorum states $\frac{a_1b_1c_1}{a_2b_2c_2} = 1$, so $a_1b_1c_1=a_2b_2c_2$.

$4S_{DEF}\leq S_{ABC}$ is equivilant to $3S_{DEF}\leq S_{ABC} - S_{DEF}$, which as shown above is equivalent to $3S_{DEF}\leq S_{CDE}+S_{FBD}+S_{AFE}$.

This can finally be rewritten as $3(a_2b_2c_2 + a_1b_1c_1) \leq (a_1b_1c_2+a_2b_2c_1)+(a_1c_1b_2+a_2c_2b_1)+(b_1c_1a_2+b_2c_2a_1)$

For the first set of parenthesis on the right, $a_1b_1c_2+a_2b_2c_1 = a_1b_1c_1(\frac{c_2}{c_1}) + a_2b_2c_2(\frac{c_1}{c_2}) = (\frac{c_1}{c_2}+\frac{c_2}{c_1})a_1b_1c_1$. Remember, $a_1b_1c_1=a_2b_2c_2$ due to Ceva's theorum.

Thus, the above inequality, with the first set of parenthesis, can be written as $2(a_1b_1c_1)\leq (\frac{c_1}{c_2}+\frac{c_2}{c_1})a_1b_1c_1$, which is true. This same method can be repeated for the other two sets of parenthesis to get the required statement.

Answer 2

Let the segment ratios be

$$x= \frac {CD}{DB}, \>\>\> y= \frac {AE}{EC}, \>\>\> z= \frac {BF}{FA}$$

According to the Ceva's theorem,

$$xyx=1\tag{1}$$

Evaluate the area ratio

$$\frac{S_{AEF}}{S_{ABC}} = \frac{ \frac 12 AE\cdot AF\sin A}{\frac 12 AB\cdot AC\sin A} =\frac{1}{(1+z)(1+\frac 1y)}=\frac{y}{(1+z)(1+y)}$$

Similarly,

$$\frac{S_{BDF}}{S_{ABC}} = \frac{z}{(1+x)(1+z)}, \>\>\>\>\> \frac{S_{CDE}}{S_{ABC}} = \frac{x}{(1+y)(1+x)}$$

Then,

$$\frac{S_{DEF}}{S_{ABC}} = 1 - \frac{S_{AEF} + S_{BDF}+S_{CDE} }{S_{ABC}} $$ $$= 1- \frac{y}{(1+z)(1+y)}-\frac{z}{(1+x)(1+z)}-\frac{x}{(1+y)(1+x)}$$

Simplify the expression,

$$\frac{S_{DEF}}{S_{ABC}} = \frac{2}{(1+x)(1+y)(1+z)} $$

Apply the AM-GM inequality to the denominator,

$$(1+x)(1+y)(1+z)=1+x+y+z+xy+yz+zx+xyz \ge 8[(xyz)^4]^{\frac 18}=8$$

Thus $$\frac{S_{DEF}}{S_{ABC}} \le \frac 14$$