Inequality with Roots, Squares and the Area

Solution 1

Rotate $\Delta CBP\;$ around $B\;$ and away from $A\;$ through $60^{\circ}$ into position $C'BP'.\;$ Observe that this creates aquilateral triangles $BCC'\;$ and $BPP'.\;$

This gives us $PB=PP',\;$ and $PC=P'C'\;$ so that

(1)

$PA+PB+PC=AP+PP'+P'C'\ge AC'.$

The Law of Cosines in $\Delta ABC'\;$ gives (with $\angle ABC=\beta )$

\displaystyle\begin{align} AC'^2 &= AB^2+BC'^2-2\cdot AB\cdot BC'\cos\angle ABC'\\ &=c^2+a^2-2ac\cos (\beta+60^{\circ})\\ &=c^2+a^2-2ac\left(\cos 60^{\circ}\cos\beta-\sin 60^{\circ}\sin\beta\right)\\ &=c^2+a^2-2ac\left(\frac{1}{2}\cos\beta-\frac{\sqrt{3}}{2}\sin\beta\right)\\ &=c^2+a^2-\left(ac\cos\beta+\sqrt{3}ac\sin\beta\right)\\ &=c^2+a^2-\frac{a^2+c^2-b^2}{2}+\sqrt{3}\cdot 2S\\ &=\frac{2a^2+2c^2-a^2-c^2+b^2}{2}+2\sqrt{3}S\\ &=\frac{a^2+c^2+b^2}{2}+2\sqrt{3}S. \end{align}

Thus $AC'=\displaystyle\sqrt{\frac{a^2+b^2+c^2}{2}+2\sqrt{3}S}.\;$ With (1), this implies

$\displaystyle PA+PB+PC \ge \displaystyle\sqrt{\frac{a^2+b^2+c^2}{2}+2\sqrt{3}S},$

which is the same as the required

$\displaystyle \sqrt{2}(PA+PB+PC)\ge\sqrt{a^2+b^2+c^2+4\sqrt{3}S}.$

For equality, we need $P,P'\in AC'.$

In such a case, $\angle BP'C'=120^{\circ},\;$ for, it's complementary to $\angle BP'P.\;$ This makes $\angle BPC=120^{\circ}.\;$ Also, $\angle BPA=120^{\circ},\;$ as complementary to $\angle BPP'.\;$ Thus, $\angle APC=120^{\circ}\;$ also, which makes $P\;$ the Fermat-Torricelli point in $\Delta ABC.$ Naturally, this argument does not work when $\angle ABC\gt 120^{\circ}.$

Solution 2

We'll consider the case in which $A,B,C\lt 120^{\circ}.\;$ Let $T\;$ be the Fermat-Toricelli point. Denote $TA=x,\;$ $TB=y\;$ and $TC=z.\;$ Choose $T=0,\;$ $A=x,\;$ $B=yu\;$ and $C=zu^2,\;$ $\displaystyle u=-\frac{1}{2}+i\frac{\sqrt{3}}{2}.\;$ We have:

\displaystyle\begin{align} a^2 &= y^2+yz+z^2,\\ b^2 &= z^2+zx+x^2,\\ a^2 &= x^2+xy+y^2,\\ 4S\sqrt{3}&=3(xy+yz+zx), \end{align}

so that $a^2+b^2+c^2+4S\sqrt{3}=2(x+y+z)^2.\;$ Thus, our inequality reduces to $PA+PB+PC\ge TA+TB+TC,\;$ which is known. Let's prove it, though.

Let $P=w.\;$ We need to show that

$|w-x|+|w-yu|+|w-zu^2|\ge x+y+z$

which is equivalent to

$|w-x|+|u^2 ||w-yu|+|u||w-zu^2|\ge x+y+z.$

But

\begin{align} |w-x|+|u^2 ||w-yu|+|u||w-zu^2 |&\ge |w(1+u^2+u)-(x+y+z)|\\ &=x+y+z. \end{align}

Naturally, equality holds iff $w=0,\;$ i.e., when $P=T.$

Generalization

If point $P\;$ lies in the interior of a convex polygon $A_1A_2\ldots A_n,\;$ $(A_{n+1}=A_1),\;$ then letting $|A_kA_{k+1}|=a_k,\;$ $|PA_k|=R_k,\;$ $k=1,2,\ldots,a_n,\;$ and $S=[A_1A_2\ldots A_n],\;$ the area of the polygon, then the following inequality holds

$\displaystyle\sum_{k=1}^{n}R_k\ge\sqrt{\frac{1}{2}\left(\sum_{k=1}^{n}a^2_k+4\tan\frac{\pi}{n}S\right)}.$

Proof

Let $\angle A_kPA_{k+1}=\theta_k.\;$ Then, by the Law of Cosines, $a^2_k=R^2_k+R^2_{k+1}-2R_kR_{k+1}\cos\theta_k\;$ and $S=\displaystyle\frac{1}{2}\sum_{k=1}^{n}R_kR_{k+1}\sin\theta_k.\;$ Thus

\displaystyle\begin{align} (RHS)^2 &= \frac{\displaystyle\sum_{k=1}^{n}a^2_k+4\tan\frac{\pi}{n}S}{2}\\ &= \frac{\displaystyle\sum_{k=1}^{n}(R^2_k+R^2_{k+1}-2R_kR_{k+1}\cos\theta_k)+2\tan\frac{\pi}{n}\sum_{k=1}^{n}R_kR_{k+1}\sin\theta_k}{2}\\ &= \sum_{k=1}^{n}R^2_k -\sum_{k=1}^{n}R_kR_{k+1}\left(\cos\theta_k-\tan\frac{\pi}{n}\sin\theta_k\right)\\ &= \sum_{k=1}^{n}R^2_k -\sum_{k=1}^{n}R_kR_{k+1}\frac{\displaystyle\cos\left(\theta_k+\frac{\pi}{n}\right)}{\cos\displaystyle\frac{\pi}{n}}\\ &= \sum_{k=1}^{n}R^2_k +\frac{\displaystyle\sum_{k=1}^{n}R_kR_{k+1}}{\cos\displaystyle\frac{\pi}{n}}\cos\left(\pi-\theta_k-\frac{\pi}{n}\right)\\ &\le\sum_{k=1}^{n}R^2_k +\frac{\displaystyle\sum_{k=1}^{n}R_kR_{k+1}}{\cos\displaystyle\frac{\pi}{n}}\\ &\le\sum_{k=1}^{n}R^2_k +2\sum_{k=1}^{n}R_kR_{k+1}\\ &=\left(\sum_{k=1}^{n}R_k\right)^2. \end{align}

Equality is only achieved for $n=3\;$ and $P=T\;$- the Fermat-Toricelli point, and all angles of the triangle less than $\displaystyle\frac{2\pi}{3}.$

Acknowledgment

The problem has been kindly communicated to me by Dan Sitaru, along with a solution (Solution 1 above;) Solution 2 is by Leo Giugiuc; Generalization and its proof are due to Marian Dinca.