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Let the sides of triangle $T$ have lengths $a$, $b$ and $c$ and the corresponding altitudes have lengths $H_a$, $H_b$ and $H_c$.
By the triangle inequality, we have $a+b > c$, $b+c > a$ and $c+a > b$ and so $a+b+c > 2c$, $a+b+c > 2a$ and $a+b+c > 2b$.
Also, since $T$ has area $1$, we have $\frac{1}{2}aH_a=1$, $\frac{1}{2}bH_b=1$ and $\frac{1}{2}cH_c=1$ and so $aH_a=2$, $bH_b=2$ and $cH_c=2$. $$M=(a+b+c)(H_a+H_b+H_c)=(a+b+c)H_a+(a+b+c)H_b+(a+b+c)H_c$$ $$ > 2aH_a+2bH_b+2cH_c=4+4+4=12$$ so $M> 12$ and hence statement E is false.
Altitude Inequalities
Age 14 to 16
ShortChallenge Level 
Let the sides of triangle $T$ have lengths $a$, $b$ and $c$ and the corresponding altitudes have lengths $H_a$, $H_b$ and $H_c$.
By the triangle inequality, we have $a+b > c$, $b+c > a$ and $c+a > b$ and so $a+b+c > 2c$, $a+b+c > 2a$ and $a+b+c > 2b$.
Also, since $T$ has area $1$, we have $\frac{1}{2}aH_a=1$, $\frac{1}{2}bH_b=1$ and $\frac{1}{2}cH_c=1$ and so $aH_a=2$, $bH_b=2$ and $cH_c=2$. $$M=(a+b+c)(H_a+H_b+H_c)=(a+b+c)H_a+(a+b+c)H_b+(a+b+c)H_c$$ $$ > 2aH_a+2bH_b+2cH_c=4+4+4=12$$ so $M> 12$ and hence statement E is false.
This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.
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