# An Inequality in Triangle with a Constraint

### Solution 1

$c^2\cdot CD+b^2\cdot BD=a\cdot AD^2+a\cdot BD\cdot CD,$

or, in our case, $2a\cdot BD\cdot CD=c^2\cdot CD+ b^2\cdot BD,$ implying $\displaystyle 2a=\frac{b^2}{CD}+\frac{c^2}{BD}.$

By Radon's inequality, $\displaystyle \frac{b^2}{CD}+\frac{c^2}{BD}\ge\frac{(b+c)^2}{CD+BD}=\frac{(b+c)^2}{a}.$ It follows that $2a^2\ge (b+c)^2,$ or else $a\sqrt{2}\ge b+c.$

### Solution 2

We fix center of coordinates as the foot $D$ and $BC$ laying on the horizontal axis. Let $AB=c,$ $AC=b,$ $BD=x,$ $CD=y,$ $AD=r=\sqrt{xy}$. Let $\theta\in (0,\pi)$ be the angle made by $AD$ with the semiaxis $x\geq 0$. By the Law of Cosines we have:

$c=\sqrt{x^2+xy+2xr\cos{\theta}}\\ b=\sqrt{y^2+xy-2yr\cos{\theta}}$

Now a key step is to notice that $(\sqrt{2}x)^2-c^2$ and $(\sqrt{2}y)^2-b^2$ have a common factor. Indeed, denote by $U=x-y-2r\cos(\theta)$.

Then $(\sqrt{2}x)^2-c^2=xU$ and $(\sqrt{2}y)^2-b^2=-yU$, so $\displaystyle \sqrt{2}x-c=\frac{xU}{\sqrt{2}x+c}$ and $\displaystyle \sqrt{2}y-b=\frac{-yU}{\sqrt{2}y+b}$.

Thus

\displaystyle \begin{align} \sqrt{2}x-c+\sqrt{2}y-b&=\frac{xU}{\sqrt{2}x+c}+\frac{-yU}{\sqrt{2}y+b}\\ &=\frac{xb-yc}{(\sqrt{2}x+c)(\sqrt{2}y+b)}U\\ &=\frac{(xb)^2-(yc)^2}{(xb+yc)(\sqrt{2}x+c)(\sqrt{2}y+b)} U\\ &=\frac{U^2(x+y)xy}{(xb+yc)(\sqrt{2}x+c)(\sqrt{2}y+b)}\geq 0. \end{align}

Hence $\sqrt{2}x-c+\sqrt{2}y-b=\sqrt{2}a-(b+c)\geq 0.$ QED

### Acknowledgment

The problem from a 2004 book by K. R. S. Sastry was posted at the Peru Geometrico facebook group by Kadir Altintas and kindly reposted by Leo Giugiuc, along with a solution (Solution 1) of his, at the CutTheKnotMath facebook page. Solution 2 is by Bogdan Lataianu.