Question

If $p_1,p_2,p_3$ are the altitudes of a triangle which circumscribed a circle of diameter $\frac{4}{3}$units, then the least vale of $p_1+p_2+p_3$ is

Answer 1

We have,

$p_1=\frac{2△}{a},p_2=\frac{2△}{b},p_3=\frac{2△}{c}$

$\therefore p_1+p_2+p_3\ge {\left(p_1{p}_2{p}_3\right)}^{\frac{1}{3}}$

$=3{\left\{\frac{{\left(2△\right)}^3}{abc}\right\}}^{\frac{1}{3}}=6△{\left(\frac{1}{abc}\right)}^{\frac{1}{3}}$

$=6rs{\left(\frac{1}{abc}\right)}^{\frac{1}{3}}=6r\frac{a+b+c}{2}{\left(\frac{1}{abc}\right)}^{\frac{1}{3}}$

$=9r\left(\frac{a+b+c}{3}\right){\left(\frac{1}{abc}\right)}^{\frac{1}{3}}\ge 9r(\frac{AM}{GM}\ge 1)$

(Equality occurs when $p_1=p_2=p_3$ and $a=b=c,$ i.e., when $△ABC$ is equilateral)

$\therefore p_1+p_2+p_3\ge 9\times \frac{2}{3}=6$