A Cyclic Inequality in Three Variables XXI

Solution 1

Note that $\displaystyle \sqrt{\frac{ab}{7}}\le\frac{a^2-ab+b^2}{\sqrt{a^2+5ab+b^2}}.\,$ Indeed, that is equivalent to $(a-b)^2(7a^2-ab+7b^2)\ge 0,\,$ which is true, with equality for $a=b.\,$

Taking the product of three such inequalities we obtain the required one. Equality when $a=b=c.$

Solution 2

Observe that

$\displaystyle a^2-ab+b^2=\frac{3}{4}(a-b)^2+\frac{1}{4}(a+b)^2\ge\frac{1}{4}(a+b)^2,\text{and}\\ \displaystyle a^2+5ab+b^2=-\frac{3}{4}(a-b)^2+\frac{7}{4}(a+b)^2\le\frac{7}{4}(a+b)^2.$

Also, for $a,b,c\gt 0,\,$ $(a+b)(b+c)(c+a)\ge 8abc.\,$ Combining that

\displaystyle \begin{align} \prod_{cycl}\frac{a^2-ab+b^2}{\sqrt{a^2+5ab+b^2}}&\ge\frac{\displaystyle \left(\frac{1}{4}\right)^3\prod_{cycl}(a+b)^2}{\displaystyle \left(\frac{\sqrt{7}}{2}\right)^3\prod_{cycl}(a+b)}\\ &=\frac{1}{56\sqrt{7}}(a+b)(+c)(c+a)\\ &\ge\frac{abc}{7\sqrt{7}}. \end{align}

Solution 3

$\displaystyle a^2-ab+b^2=\frac{3}{4}(a-b)^2+\frac{1}{4}(a+b)^2\ge\frac{1}{4}(a+b)^2.\,$ Similarly, $\displaystyle b^2-bc+c^2\ge\frac{(b+c)^2}{4}\,$ and $\displaystyle c^2-ca+a^2\ge\frac{(c+a)^2}{4}.\,$ Thus,

\displaystyle \begin{align} \prod_{cycl}(a^2-ab+b^2) &\ge \frac{(a+b)^2(b+c)^2(c+a)^2}{64}\\ &=\frac{\displaystyle \prod_{cycl}(a+b)\cdot\prod_{cycl}(a+b)}{64}\\ &\ge\frac{8abc}{64}(a+b)(b+c)(c+a)\\ &=\frac{(a+b)(b+c)(c+a)}{8}. \end{align}

Suffice it to prove that

$\displaystyle \frac{\displaystyle abc\prod_{cycl}(a+b)}{8}\ge\frac{abc}{7\sqrt{7}}\prod_{cycl}\sqrt{a^2+5ab+b^2}$

which is the same as

(a)

$\displaystyle \prod_{cycl}\frac{\sqrt{7}(a+b)}{2}\ge \prod_{cycl}\sqrt{a^2+5ab+b^2}$

Suffice it to prove that $\displaystyle \frac{\sqrt{7}(a+b)}{2}\ge\sqrt{a^2+5ab+b^2},\,$ which is equivalent to

\displaystyle \begin{align} &7(a^2+b^2+2ab)\ge 4(a^2+5ab+b^2)\,\Longleftrightarrow\\ &3(a-b)^2\ge 0, \end{align}

which is true.

Solution 4

The required inequality is equivalent to

$\displaystyle 7^3\prod_{cycl}(a^2-ab+b^2)^2\ge (abc)^2\prod_{cycl}(a^2+5ab+b^2).$

Now,

\begin{align} 7(a^2-ab+b^2)^2&-\,ab(a^2+5ab+b^2)\\ &=7(a^2+b^2)^2-15ab(a^2+b^2)+2(ab)^2\\ &=(a-b)^2(7a^2+7b^2-ab)\ge 0, \end{align}

so $7(a^2-ab+b^2)^2\ge ab(a^2+5ab+b^2).\,$ Rotating $a,b,c\,$ and taking the product yields the required inequality.

Solution 5

Consider

\begin{align} 7(a^2-ab+b^2)^2&-\,ab(a^2+5ab+b^2)\\ &=7(a^2-ab+b^2)^2-\,ab(a^2-ab+b^2)-6a^2b^2\\ &=7(a^2-ab+b^2)^2-\,7ab(a^2-ab+b^2)\\ &\qquad\qquad\qquad\qquad+\,6ab(a^2-ab+b^2)-6a^2b^2\\ &=7(a^2-ab+b^2)(a^2-ab+b^2-ab)\\ &\qquad\qquad\qquad\qquad+\,6ab(a^2-ab+b^2-ab)\\ &=(a-b)^2[(a^2-ab+b^2)+6(a^2+b^2)]\ge 0. \end{align}

Hence, $\displaystyle \frac{a^2-ab+b^2}{\sqrt{a^2+5ab+b^2}}\ge\frac{\sqrt{ab}}{\sqrt{7}}.\,$ Rotating $a,b,c\,$ and taking the product yields the required inequality.

Solution 6

First note that $\displaystyle \frac{x^2-x+1}{\sqrt{x^2+5x+1}}\ge\frac{1}{2\sqrt{7}}(x+1),\,$ $(xyz=1)\,$ the RHS being the tangent of the LHS:

With this in mind, set $\displaystyle \frac{a}{b}=x,\,$ $\displaystyle \frac{b}{c}=y,\,$ $\displaystyle \frac{c}{a}=z.\,$ The inequality reduces to

$\displaystyle \frac{x^2-x+1}{\sqrt{x^+5x+1}}\ge\frac{1}{7\sqrt{7}}.$

We have

\displaystyle\begin{align} LHS &\ge \frac{x+1}{2\sqrt{7}}\cdot\frac{y+1}{2\sqrt{7}}\cdot\frac{z+1}{2\sqrt{7}}\\ &=\frac{(x+1)(y+1)(z+1)}{8\cdot 7\sqrt{7}}\\ &\ge\frac{2\sqrt{x}\cdot 2\sqrt{y}\cdot 2\sqrt{z}}{8\cdot 7\sqrt{7}}\\ &=\frac{1}{7\sqrt{7}}. \end{align}

Solution 7

Let's find $k\gt \,$ for which $k(a^2+5ab+b^2)\le a^2+b^2,\,$ i.e.,

$(k-1)a^2-5ab+(k-1)b^2\ge 0.$

The discriminant of the quadratic form is $5^2-4(k-1)^2,\,$ which is not positive for $2|k-1|\le 5.\,$ The simplest suitable value is $\displaystyle k=\frac{7}{2}.$

Now we have

\displaystyle \begin{align} &\prod_{cycl}\sqrt{a^2+5ab+b^2}\le\left(\frac{7}{2}\right)^{\frac{3}{2}}\prod_{cycl}\sqrt{a^2+b^2}\\ &\prod_{cycl}(a^2-ab+b^2)\ge\prod_{cycl}\left(\frac{a^2+b^2}{2}\right). \end{align}

Dividing one by the other and taking into account that $a^2+b^2\ge 2ab,\,$ we obtain the required inequality.

Acknowledgment

This problem from the Romanian Mathematical Magazine, has been kindly posted at the CutTheKnotMath facebook page by Dan Sitaru. Solution 1 is by Leo Giugiuc; Solution 2 is by Kevin Soto Palacios; Solution 3 by Soumava Chakraborty; Solution 4 by Diego Alvariz; Solution 5 by Ravi Prakash; Solution 6 is by Sladjan Stankovik; Solution 7 is by Srinivas Vemuri.