# Leo Giugiuc's Second Lemma And Applications

### Proof 1

\begin{align} 2(m_a+m_b) &= |\overrightarrow{AB}+\overrightarrow{AC}|\lt |\overrightarrow{AB}|+|\overrightarrow{AC}|+|\overrightarrow{BA}|+|\overrightarrow{BC}|\\ &= a+b+2c < (a+b+2c)+2(a+b-c)\\ &= 3(a+b). \end{align}

### Proof 2

The required inequality is equivalent to $\displaystyle\frac{2}{3}(m_a+m_b)\le a+b.\,$ Geometrically, this becomes $AG+BG\le AC+BC.$

Let $A',B\,$ be the midpoints of $BC\,$ and $AC,\,$ respectively. In $\Delta BGA',\,$ $BG\lt GA'+A'B\,$ so that $AG+BG\lt AG+GA'+A'B=AA'+BB'.$

In $\Delta AA'C,\,$ $AA'\lt A'C'+AC\,$ so that

$AG+BG\lt AA'+BB'\lt A'C+AC+A'B=BC+AC.$

### Proof 3

Let, for a point $X,\,$ $x\,$ denotes the corresponding complex numbers. $\displaystyle g=\frac{a+b+c}{3}.$

\displaystyle\begin{align} AG+BG &= \left|a-\frac{a+b+c}{3}\right|+\left|b-\frac{a+b+c}{3}\right|\\ &=\left|\frac{a-b+a-c}{3}\right|+\left|\frac{b-a+b-c}{3}\right|\\ &\le\frac{|a-b|+|a-c|}{3}+\frac{|b-a|+|b-c|}{3}\\ &=\frac{2\cdot AB+AC+BC}{3}\\ &=\frac{2\cdot (AC+BC)+(AC+BC)}{3}\\ &\le AC+BC. \end{align}

### Proof 4

This is just a follow-up on the starting point of Proof 2.

The required inequality is equivalent to $AG+BG\lt AC+BC,\,$ i.e. $AB+AG+BG\lt AB+AC+BC.\,$ But, in general, given two convex polygons, one within the other, the perimeter of the outer polygon exceeds the permiter of the inner one. The lemma then follows from the fact that $\Delta ABG\,$ is within $\Delta ABC.\,$

### Proof 5

Lemma is a special case of the following assertion:

If $M\,$ is an interior point of $\Delta ABC,\,$ then $MB + MC < AB + AC.$

To prove that, consider the ellipse whose focii are $B\,$ and $C\,$ and the major axis is $AB + AC.\,$ Point $M\,$ being inside $\Delta ABC,\,$ it is also inside the ellipse, so that $MB + MC < AB + AC.$

### Proof 6

This is another proof of the generalization frm Proof 5.

There are $x,y,z\gt 0\,$ such that $\displaystyle\overrightarrow{XM}=\frac{x\overrightarrow{XA}+y\overrightarrow{XB}+z\overrightarrow{XC}}{x+y+z},\,$ for any $X\,$ in the plane of the triangle. Taking $X=B,\,$ $\overrightarrow{BM}=\displaystyle\frac{x\overrightarrow{BA}+z\overrightarrow{BC}}{x+y+z};\,$ and taking $X=C,\,$ $\displaystyle\overrightarrow{CM}=\frac{x\overrightarrow{CA}+y\overrightarrow{CB}}{x+y+z}.\,$ We have

$\displaystyle MB=|\overrightarrow{BM}|-\left|\frac{x\overrightarrow{BA}+z\overrightarrow{BC}}{x+y+z}\right|\lt \frac{xc+xa}{x+y+z}.$

Similarly $\displaystyle MC\lt\frac{xb+ya}{x+y+z},\,$ such that $\displaystyle MB+MC\lt\frac{(y+z)a+xb+xc}{x+y+z}.\,$ Suffice it to show that $\displaystyle\frac{(y+z)a+xb+xc}{x+y+z}\le b+c.\,$ But this is equivalent to $(y+z)(b+c-a)\ge 0\,$ which is true.

### Proof 7

In $\Delta AMC,\,$ $AM\lt AC+MC,\,$ i.e., $m_a\lt b+\displaystyle\frac{a}{2}.\,$ In $\Delta BNC,\,$ $BN\lt BC+NC,\,$ i.e., $m_b\lt a+\displaystyle\frac{b}{2}.\,$ Summing up, $\displaystyle m_a+m_b\lt\frac{3(a+b)}{2},\,$ which is equivalent to the required inequality.

### Applications

Prove that in any $\Delta ABC,$

\begin{align}\min (a,b,c)&+2\max(m_a,m_b,m_c)\\&\ge\max(a,b,c)+2\min(m_a,m_b,m_c). \end{align}

Solution

Let WLOG, $a\ge c\ge b.\,$ Then $m_b\ge m_c\ge m_a.\,$ Suffice it to show that $2(m_b-m_a)\ge a-b,\,$ which is the same as

$\sqrt{2a^2-b^2+2c^2}-\sqrt{-a^2+2b^2+2c^2}\ge a-b,$

or,

$\displaystyle\frac{(2a^2-b^2+2c^2)-(-a^2+2b^2+2c^2)}{\sqrt{2a^2-b^2+2c^2}+\sqrt{-a^2+2b^2+2c^2}}\ge a-b$

which, in turn, is equivalent to

$\displaystyle (a-b)\frac{3(a+b)}{2(m_a+m_b)}\ge a-b.$

But $a-b\ge 0\,$ and from the lemma $\displaystyle\frac{3(a+b)}{2(m_a+m_b)}\gt 1$ which confirms the above unequality.

### Acknowledgment

The above lemma has been kindly posted by Leo Giugiuc at the CuTheKnotMath facebook page. Leo has later communicated to me in a private email the proof and an application - his solution to a problem from the awesomemath.org collection (Problem S391 by Titu Andreescu). Proofs 2 and 3 are by Marian Dinca. Proofs 5 and 6 are by Leo Giugiuc; Proof 7 is by Gabriela Negutescu.