An Inequality in Triangle with Medians, Sides and Circumradius

Problem

An Inequality in Triangle with Medians, Sides and Circumradius

Proof 1

It's known that $\displaystyle m_a^2=\frac{b^2+c^2}{2}-\frac{a^2}{4}.\,$ So, using the Law of Cosines and the addition formula for cosines, we have

$\displaystyle\begin{align} m_a^2 &= \frac{2(b^2+c^2)-a^2}{4}=\frac{2(b^2+c^2)-(b^2+c^2-2bc\cos A)}{4}\\ &=\frac{b^2+c^2+2bc\cos A}{4}=\frac{b^2+c^2-2bc\cos (B+C)}{4}\\ &=\frac{(b\cos B-c\cos C)^2+(b\sin B+c\sin C)^2}{4}\\ &\ge\frac{(b\sin B+c\sin C)^2}{4}, \end{align}$

so that $\displaystyle m_a\ge\frac{b\sin B+c\sin C}{2}=\frac{b^2+c^2}{4R},\,$ by the Law of Sines.

Proof 2

We choose $A=(0,1),\,$ $B=(-k, 0)\,)$ and $C=(t,0),\,$ $k=\cos B\,$ and $t=\cos C.\,$ We have

$2R=\sqrt{(k^2+1)(t^2+1)},\,$ $2m_a=\sqrt{(t-k)^2+4},\,$ $b=\sqrt{t^2+1},\,$ $c=\sqrt{k^2+1}.\,$ We need to show that

$\begin{align} &\sqrt{(k^2+1)(t^2+1)}\cdot \sqrt{(t-k)^2+4}\ge k^2+t^2+2\;\Longleftrightarrow\\ &(k^2+1)(t^2+1)[(t-k)^2+4]\ge (k^2+t^2+2)^2\;\Longleftrightarrow\\ &(k^2+1)(t^2+1)(t-k)^2\ge (k^2+t^2)^2+4(k^2+t^2+1)\\ &\qquad\qquad\qquad\qquad\qquad\qquad-4(k^2t^2+k^2+t^2+1)\;\Longleftrightarrow\\ &(k^2+1)(t^2+1)(t-k)^2\ge (k^2+t^2)^2-4k^2t^2\;\Longleftrightarrow\\ &(k^2+1)(t^2+1)(t-k)^2\ge (t-k)^2(t+k)^2\;\Longleftrightarrow\\ &(t-k)^2[(k^2+1)(t^2+1) -(t+k)^2]\;\Longleftrightarrow\\ &(t-k)^2(1-kt)^2\ge 0, \end{align}$

which is true. Obviously, equality holds if either $k=t\,$ or $kt=1,\,$ i.e., if $b=c\,$ or $\angle A=90^{\circ}.$

Proof 3

For a symmedian, $s_a=\displaystyle \frac{2bcm_a}{b^2+c^2}.\,$ For an altitude, $\displaystyle h_a=\frac{bc}{2R}.\,$ Since $s_a\ge h_a,\,$ $\displaystyle \frac{2bcm_a}{b^2+c^2}\ge\frac{bc}{2R},\,$ or, equivalently, $4Rm_a\ge b^2+c^2.$

Acknowledgment

The proof has been generously shared on facebook by Marian Dincă. Proof 2 is be Leo Giugiuc who informed us that the inequality is known as Tereshin's. Proof 3 is by Adil Abdullayev.

 

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