# Dan Sitaru's Cyclic Inequality in Three Variables III

### Solution 1

The problem is easily equivalent to

$\displaystyle 4\sqrt{\sum_{cycl}\frac{a+1}{a^2}}\ge\sqrt{6}(7-a-b-c),$

for $a,b,c\gt 0.$

Let's find $m,n\,$ such that $\displaystyle \frac{\sqrt{x+1}}{x}\ge mx+n,\,$ for $x\gt 0.\,$ This is the same as $f(x)=mx^2+nx-\sqrt{x+1}\le 0.\,$ Assuming $f(1)=0,\,$ $m+n=\sqrt{2}.\,$ Assuming $f'(1)=0,\,$ $\displaystyle 2m+n=\sqrt{1}{2\sqrt{2}},\,$ solving which $\displaystyle m=-\frac{3}{2\sqrt{2}}\,$ and $\displaystyle n=\frac{7}{2\sqrt{2}}.$

We shall prove that indeed $g\frac{\sqrt{x+1}}{x}\ge\frac{-3x+7}{2\sqrt{2}},\,$ for $x\gt 0.$

If $\displaystyle 0\lt x\le\frac{7}{3},\,$ the inequality is equivalent to $(x-1)^2(9x^2-24x-8)\le 0,\,$ which is true because $\displaystyle \frac{4+2\sqrt{6}}{3}|gt\frac{7}{3}.\,$ If $\displaystyle g\gt\frac{7}{3}\,$ then, obviously, $\displaystyle \frac{\sqrt{x+1}}{x}\gt 0\gt\frac{-3x+7}{2\sqrt{2}}.$

Back to the original problem, by Jensen's inequality,

\displaystyle\begin{align} \sqrt{3\sum_{cycl}\frac{a+1}{a^2}}&\ge\sum_{cycl}\frac{\sqrt{a+1}}{a}\\ &\ge\frac{3(7-a-b-c)}{2\sqrt{2}}. \end{align}

Multiply by $4\,$ to get the required inequality.

### Solution 2

This solution is more of an illustration to the first one.

$\displaystyle \sqrt{3\sum_{cycl}\frac{a+1}{a^2}}\ge\sum_{cycl}\frac{\sqrt{a+1}}{a}.$

Thus, suffice it to prove that, for $x\gt 0,\,$ $\displaystyle \frac{\sqrt{x+1}}{x}\gt\frac{7-3x}{2\sqrt{2}}.\,$ We can see that $\displaystyle \left(\frac{\sqrt{x+1}}{x}\right)(1)=\sqrt{2}\,$ and $\displaystyle \left(\frac{\sqrt{x+1}}{x}\right)'(1)=-\frac{3}{2\sqrt{2}}\,$ which makes $\displaystyle \frac{-3x+7}{2\sqrt{2}}\,$ tangent to $\displaystyle \frac{\sqrt{x+1}}{x}\,$ at $x=1.$

The second derivative of $\displaystyle \frac{\sqrt{x+1}}{x}\,$ is easily seen to be negative, making the function convex and insuring the inequality

$\displaystyle \frac{\sqrt{x+1}}{x}\gt\frac{7-3x}{2\sqrt{2}}.$

### Extra

Once we surmised the secret behind the problem design, we may try to modify the problem. For example, at $x=3,\,$ the tangent to function $\displaystyle f(x)=\frac{\sqrt{x+1}}{x}\,$ is given by $\displaystyle y=-\frac{5}{36}x+\frac{13}{12},\,$ implying

\displaystyle\begin{align}\sqrt{\sum_{cycl}\frac{a+1}{a^2}}&\ge\sum_{cycl}\frac{\sqrt{a+1}}{a\sqrt{3}}\\ &\ge \frac{\sqrt{3}}{108}[117-5(a+b+c)]. \end{align}

### Solution 3

Let $x=a-1$, $y=b-1$, and $z=c-1$. Thus, $x,y,z\gt 0.\,$ Let us rewrite the inequality as

$\displaystyle 4\sqrt{\frac{x+1}{6x^2}+\frac{y+1}{6y^2}+\frac{z+1}{6z^2}}+(x+y+z)\geq 7.$

$x+y+z\geq3(xyz)^{1/3}~\text{(AM-GM)}$

and

\displaystyle \begin{align} &4\sqrt{\frac{x+1}{6x^2}+\frac{y+1}{6y^2}+\frac{z+1}{6z^2}}\\ &=4\sqrt{\frac{1}{3}\left[\frac{1}{2}\left(\frac{1}{x}+\frac{1}{x^2}\right)+ \frac{1}{2}\left(\frac{1}{y}+\frac{1}{y^2}\right) +\frac{1}{2}\left(\frac{1}{z}+\frac{1}{z^2}\right)\right]} \\ &\geq 4\sqrt{\frac{1}{3}\left(\frac{1}{x^{3/2}}+\frac{1}{y^{3/2}}+\frac{1}{z^{3/2}}\right)}~\text{(bracket-wise AM-GM)} \\ &\geq\frac{4}{3}\left(\frac{1}{x^{3/4}}+\frac{1}{y^{3/4}}+\frac{1}{z^{3/4}}\right)~\text{(Jensen's/concavity of the square root)} \\ &\geq 4\left(\frac{1}{xyz}\right)^{1/4}.~\text{(AM-GM)} \end{align}

Let $q=(xyz)^{1/7}$. Thus,

\displaystyle \begin{align} LHS&\geq 4\left(\frac{1}{q}\right)^{7/4}+3q^{7/3} \\ &=7\left[\frac{\left(\frac{1}{q}\right)^{7/4}}{7/4}+\frac{q^{7/3}}{7/3}\right]\\ &\geq 7~\text{(Young's inequality)}. \end{align}

### Solution 3'

Let $x=a-1$, $y=b-1$, and $z=c-1$. Thus, $x,y,z\gt 0.\,$ Let us rewrite the inequality as

$\displaystyle 4\sqrt{\frac{x+1}{6x^2}+\frac{y+1}{6y^2}+\frac{z+1}{6z^2}}+(x+y+z)\geq 7.$

$x+y+z\geq3(xyz)^{1/3}~\text{(AM-GM)}$

and

\displaystyle \begin{align} &4\sqrt{\frac{x+1}{6x^2}+\frac{y+1}{6y^2}+\frac{z+1}{6z^2}} \\ &=4\sqrt{\frac{1}{6}\left(\frac{1}{x}+\frac{1}{x^2}+\frac{1}{y}+\frac{1}{y^2} +\frac{1}{z}+\frac{1}{z^2}\right)} \geq 4\left(\frac{1}{xyz}\right)^{1/4}.~\text{(AM-GM)} \end{align}

Let $q=(xyz)^{1/7}$. Thus,

$\displaystyle LHS\geq4\left(\frac{1}{q}\right)^{7/4}+3q^{7/3}=7\left[\frac{\left(\frac{1}{q}\right)^{7/4}}{7/4}+\frac{q^{7/3}}{7/3}\right] \geq 7~\text{(Young's inequality)}.$

### Solution 4

Proof by progressive reduction to one single variable.

Let $\displaystyle f=4 \sqrt{\frac{a}{(a-1)^2}+\frac{b}{(b-1)^2}+\frac{c}{(c-1)^2}}-\sqrt{6} (-a-b-c+10).\,$ Let $x = \frac{1}{( a-1)},\,$ $y = \frac{1}{( b-1)},\,$ $z = \frac{1}{( c-1)}.\,$ Now,

\displaystyle\begin{align} f&=4 \sqrt{x^2+x+y^2+y+z^2+z}+\sqrt{6}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}- 7\right)\\ &\geq 4 \sqrt{x^2+x+y^2+y+z^2+z}+3 \sqrt{6} \sqrt[3]{\frac{1}{x y z}}-7 \sqrt{6}\\ &\geq \sqrt{6} \left(4 \sqrt[4]{x} \sqrt[4]{y} \sqrt[4]{z}+\frac{3 }{\sqrt[3]{x} \sqrt[3]{y} \sqrt[3]{z}}-7\right) \end{align}

Now let $X=x y z$, $X\gt 0$. We can to prove that:

$\displaystyle \frac{4 X^{7/12}+3}{\sqrt[3]{X}}-7 \geq 0$

since it is a single function with one variable and its minimum is $0$ for $X=x=y=z=1$.

### Acknowledgment

Dan Sitaru has kindly posted the problem of his from the Romanian Mathematical Magazine at the CutTheKnotMath facebook page. Leo Giugiuc has commented with his solution (Solution 1). Solution 3 (and its concise version) is by Amit Itagi; Solution 4 is by N. N. Taleb.