# Problem 6 from Dan Sitaru's Algebraic Phenomenon

### Solution 1

\displaystyle \sqrt{(y+1)\cdot\frac{1}{3}}\le\frac{\displaystyle (y+1)+\frac{1}{3}}{2}=\frac{3y+4}{6}\\ \sqrt{y+1}\le\frac{\sqrt{3}}{6}(3y+4)\\ x\sqrt{y+1}\le\frac{\sqrt{3}}{6}(3xy+4x)\\ \begin{align} \sum_{cycl}x\sqrt{y+1}&\le\sum_{cycl}\frac{\sqrt{3}}{6}(3xy+4x)\le\frac{\sqrt{3}}{6}\left(3\sum_{cycl}xy+4\sum_{cycl}x\right)\\ &\le\frac{\sqrt{3}}{6}\left(3\sum_{cycl}xy+4\sum_{cycl}x\right)\le\frac{\sqrt{3}}{6}\left(\left(\sum_{cycl}x\right)^2+4\cdot 2\right)\\ &=\frac{\sqrt{3}}{6}(4+8)=2\sqrt{3}. \end{align}

### Solution 2

By the Cauchy-Schwarz inequality,

\displaystyle \begin{align} \left(\sum_{cycl}x\sqrt{y+1}\right)^2&\le\left(\sum_{cycl}x\right)\left(\sum_{cycl}x(y+1)\right)\\ &=2\left(\sum_{cycl}xy+2\right)\le 2\left(\frac{1}{3}(x+y+z)^2+2\right)\\ &=2\left(\frac{4}{3}+2\right)=2\cdot \frac{10}{3}=\frac{20}{3}. \end{align}

It follows that $\displaystyle \sum_{cycl}x\sqrt{y+1}\le\sqrt{\frac{20}{3}}\lt 2\sqrt{3}.$

### Solution 3

We first apply the Rearrangement and then the Cauchy-Schwarz inequality:

\displaystyle \begin{align} \left(\sum_{cycl}x\sqrt{y+1}\right)^2&\le\left(\sum_{cycl}x\sqrt{x+1}\right)^2\\ &=\left(\sum_{cycl}x\right)\left(\sum_{cycl}x(x+1)\right)\le 2\left(\sum_{cycl}x^2+\sum_{cycl}x\right)\\ &\le 2\left((x+y+z)^2+2\right)\\ &=2(4+2)=12. \end{align}

It follows that $\displaystyle \sum_{cycl}x\sqrt{y+1}\le 2\sqrt{3}.$

### Solution 4

Note that $x,y,z \leq 2$, so

\displaystyle \begin{align} \sum_{cycl} x^2(y+1) &= x^2y+y^2z + z^2x + x^2 + y^2 + z^2\\ &\leq 2xy+2yz+2zx + x^2+y^2+z^2\\ &= (x+y+z)^2\\ &= 4 \end{align}

From the above, by Jensen's inequality,

$\displaystyle \sum_{cycl}\sqrt{x^2(y+1)} \leq \sqrt{3}\sqrt{\sum_{cycl}x^2(y+1)} \leq \sqrt{3}\cdot 2$

### Solution 5

Let $AD=d$, $BD=m$ and $DC=n$. By Stewart's theorem,

$b^2m+c^2n=a(d^2+mn)=2amn=2mn(m+n).$

By the Cauchy-Schwarz inequality,

\displaystyle \begin{align} \sqrt{b^2m+c^2n}\sqrt{\frac{1}{m}+\frac{1}{n}}&\geq (b+c) \\ \sqrt{2mn(m+n)}\sqrt{\frac{1}{m}+\frac{1}{n}}&\geq (b+c) \\ (m+n)\sqrt{2}=a\sqrt{2}&\geq (b+c). \end{align}

### Acknowledgment

Dan Sitaru has kindly presented me his new book Algebraic Phenomenon. The above is problem 6 and its solution (Solution 1) from the book. Solution 4 is by Long Huynh Huu; Solution 5 is by Amit Itagi.