Solving triangle altitudes inequality using AM-GM

Question

If there is a triangle with sides $a \gt b \gt c$, prove that $a + v_a \ge b + v_b \ge c + v_c$ ($v_c$ is a triangle altitude perpendicular to side $c$ etc.). I know how to solve this, but the task is to prove it using AM-GM inequality. Any help would be greatly appreciated.

Answer 1

I will denote the triangle hights by $h_a,h_b,h_c$ then we have $$a-b\geq \frac{2S}{b}-\frac{2S}{a}$$ this is equivalent to $$S\le \frac{ab}{2}$$ which is true, since $$S=\frac{ab}{2}\sin(\gamma)\le \frac{ab}{2}$$

Answer 2

You can use a more general version of the AM-GM Inequality in this problem (but it is an odd restriction, as the solution is a bit unpleasant). The main step is to observe that $a\,v_a=b\,v_b=c\,v_c$ (which is twice the area of the triangle).

Suppose that $x$, $y$, $x'$, and $y'$ are positive real numbers such that $xy=x'y'$. I claim that, if $x\geq x'$, then $x+y\geq x'+y'$. The equality holds if and only if $x=x'$ (whence $y=y'$).

First, write $m:=xy=x'y'$. Then, using the Weighted AM-GM Inequality yields $$\begin{align}x+y &=\left(x'\right)^2\cdot\left(\frac{x}{(x')^2}\right)+m\cdot\left(\frac{y}{m}\right)\geq \left((x')^2+m\right)\left(\left(\frac{x}{(x')^2}\right)^{(x')^2}\,\left(\frac{y}{m}\right)^m\right)^{\frac1{(x')^2+m}} \\ &=\big((x')^2+m\big)\,\left(\frac{x^{(x')^2-m}}{(x')^{2(x')^2}}\left(\frac{(xy)^m}{m^m}\right)\right)^{\frac{1}{(x')^2+m}}=\big((x')^2+m\big)\,\left(\frac{x^{(x')^2-m}}{(x')^{2(x')^2}}\left(\frac{m^m}{m^m}\right)\right)^{\frac{1}{(x')^2+m}} \\ &=\big((x')^2+m\big)\,\left(\frac{x^{(x')^2-m}}{(x')^{2(x')^2}}\right)^{\frac{1}{(x')^2+m}}\geq\big((x')^2+m\big)\,\left(\frac{(x')^{(x')^2-m}}{(x')^{2(x')^2}}\right)^{\frac{1}{(x')^2+m}} \\ &=\frac{(x')^2+m}{\left((x')^{(x')^2+m}\right)^{\frac{1}{(x')^2+m}}}=\frac{(x')^2+m}{x'}=x'+\frac{m}{x'}=x'+y'\,. \end{align}$$