Show that $2\mathrm{AB}+2\mathrm{BC}+2\mathrm{CA}>4\mathrm{AT}+3\mathrm{BT}+2\mathrm{CT}$.

Question

Inside a triangle $\triangle\mathrm{ABC}$ we consider the point $\mathrm{T}$ which has the property that $$\angle\mathrm{BTC} = \angle\mathrm{CTA} = \angle\mathrm{ATB}.$$

Show that: $$2\mathrm{AB}+2\mathrm{BC}+2\mathrm{CA}>4\mathrm{AT}+3\mathrm{BT}+2\mathrm{CT}$$

Okey, so I am going to attach a photo where there are all my ideas. I just arrive at the point that

$$ 2\mathrm{AB} + 2\mathrm{BC} + 2\mathrm{CA} < 4\mathrm{AT} + 4\mathrm{BT} + 4\mathrm{CT}. $$

I hope my idea can help you think of a solution. I don't know what to do alone and hope you can help me. Any ideas are welcome.

\begin{gather} \angle\mathrm{BTC} = \angle\mathrm{CTA} = \angle\mathrm{ATB} = \frac{360^{\circ}}{3} = 120^{\circ}. \\ \mathrm{AB} + \mathrm{BC} + \mathrm{CA} = P(\triangle\mathrm{ABC}) \\ \mathrm{AT} + \mathrm{BT} + \mathrm{CT} \\ \end{gather}

For $\text{BAT}$ to be a $\triangle$, it must be

\begin{align*} \textsf{REPEAT} &\quad\implies\quad \begin{cases} \mathrm{AT} + \mathrm{BT} > \mathrm{AB} \\ \mathrm{BT} + \mathrm{CT} > \mathrm{BC} \\ \mathrm{CT} + \mathrm{AT} > \mathrm{CA} \end{cases} \\[0.5em] &\quad\implies\quad 2\mathrm{AT} + 2\mathrm{BT} + 2\mathrm{CT} > \mathrm{AB} + \mathrm{BC} + \mathrm{CA} \\[0.5em] &\quad\implies\quad 4\mathrm{AT} + 4\mathrm{BT} + 4\mathrm{CT} > 2\mathrm{AB} + 2\mathrm{BC} + 2\mathrm{CA}. \end{align*} \begin{gather} \mathrm{AB} > \mathrm{AT} \\ \mathrm{AB} > \mathrm{BT} \end{gather}

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Answer 1

Say $AT = a$, $BT=b$ and $CT =c$ and $AB=x$, $BC=y$ and $CA=z$. Then by the cosine theorem and special case of Cauchy inequality: $$(k^2+m^2)(1+1)\geq (k+m)^2$$we have $$x^2 = a^2+b^2 +ab = (b+{a\over 2})^2 +{3a^2\over 4} \geq {1\over 2}(b+{a\over 2} +{a\sqrt{3}\over 2})^2$$

and thus $$x\geq {1\over \sqrt{2}} (b+{a\over 2} +{a\sqrt{3}\over 2})$$ similary $$z\geq {1\over \sqrt{2}} (c+{a\over 2} +{a\sqrt{3}\over 2})$$ and $$y\geq {1\over \sqrt{2}} (c+{b\over 2} +{b\sqrt{3}\over 2})$$

Now the double sum of all expression on the right is greater than $4a+3b+2c$ (this is easy to verify) and you are done.

Answer 2

The point $T$ is called the Torricelli point or Fermat–Torricelli point and there is a connected construction of a equilateral triangle (you can also see it in the wiki page). Denote the sides of the initial triangle by $a,b,c$ and the segments $TA, TB$ and $TC$ by $x,y$ and $z$. Using the construction with equilateral triangle you can show the inequalities $$ c + a > x + y + z,\qquad b + c > x + y + z. $$ Now it is left to show $b + a > 2x + y$. WLOG, we can assume that $z\le x$ and $z\le y$ In this case $\angle C \le 60^{\circ}$ and again via the equilateral triangle, you can show $(x + y) + x < 2c \le a + b$.