Powers and Fractions Inequality

Proof 1

By the AM-GM inequality,

\displaystyle\begin{align} \frac{a^3c^3}{b^5}+\frac{ba}{c}+\frac{cb}{a} &\ge 3\left(\frac{a^3c^3}{b^5}\cdot\frac{ba}{c}\cdot\frac{cb}{a}\right)^{1/3}\\ &=\frac{3ac}{b}. \end{align}

Similarly, $\displaystyle\frac{b^3a^3}{c^5}+\frac{ac}{b}+\frac{cb}{a} \ge\frac{3ba}{c}\;$ and $\displaystyle\frac{c^3b^3}{a^5}+\frac{ac}{b}+\frac{ba}{c} \ge\frac{3cb}{a}.\;$ Adding up,

$\displaystyle\sum_{cycl}\frac{a^3c^3}{b^5}+2\sum_{cycl}\frac{ba}{c}\ge 3\sum_{cycl}\frac{ba}{c},$

which directly proves the required inequality.

Proof 2

By the AM-GM inequality,

\displaystyle\begin{align} 2\sum_{cycl}\frac{a^3c^3}{b^5}&=\left(\frac{a^3c^3}{b^5}+\frac{b^3a^3}{c^5}\right)+\left(\frac{b^3a^3}{c^5}+\frac{c^3b^3}{a^5}\right)+\left(\frac{c^3b^3}{a^5}+\frac{a^3c^3}{b^5}\right)\\ &\ge 2\frac{a^3}{bc}+2\frac{b^3}{ca}+2\frac{c^3}{ab}\\ &= \left(\frac{a^3}{bc}+\frac{b^3}{ca}\right)+\left(\frac{b^3}{ca}+\frac{c^3}{ab}\right)+\left(\frac{c^3}{ab}+\frac{a^3}{bc}\right)\\ &\ge 2\frac{ab}{c}+2\frac{bc}{a}+2\frac{ca}{b}\\ &=2\sum_{cycl}\frac{ac}{b}. \end{align}

Proof 3

The given inequality is equivalent to $\displaystyle\sum_{cycl}a^8b^8\ge a^4b^4c^4\sum_{cycl}a^2b^2.$

Let $a^2b^2=x,\;$ $b^2c^2=y,\;$ $c^2a^2=z,\;$ $x,y,z\gt 0.\;$ We need to prove that

(1)

$x^4+y^4+z^4\ge xyz(x+y+z).$

Schur's inequality for $t=2\;$ gives

(a)

$x^4+y^4+z^4+xyz(x+y+z)\ge xy(x^2+y^2)+yz(y^2+z^2)+zx(z^2+x^2).$

Now, $x^2+y^2\ge 2xy,\;$ etc., so that

(2)

$xy(x^2+y^2)+yz(y^2+z^2)+zx(z^2+x^2)\ge 2(x^2y^2+y^2z^2+z^2x^2).$

Further,

\begin{align} 2(x^2y^2+y^2z^2+z^2x^2)&-2xyz(x+y+z)\\&=(xy-yz)^2+(yz-zx)^2+(zx-xy)^2\ge 0, \end{align}

implying

(3)

$2(x^2y^2+y^2z^2+z^2x^2)\ge 2xyz(x+y+z).$

(2) and (3) give

(b)

$xy(x^2+y^2)+yz(y^2+z^2)+zx(z^2+x^2)\ge 2xyz(x+y+z).$

(a) and (b) add up to the required (1).

Proof 4

\displaystyle\begin{align} 4\sum_{cycl}\frac{a^3b^3}{c^5} &= \sum_{cycl}\left(2\frac{a^3b^3}{c^5} + \frac{b^3c^3}{a^5} + \frac{c^3a^3}{b^5}\right)\\ &\ge \sum_{cycl}4\left(\frac{a^3b^3}{c^5}\cdot\frac{a^3b^3}{c^5}\cdot \frac{b^3c^3}{a^5}\cdot \frac{c^3a^3}{b^5}\right)^{1/4}\\ &= 4\sum_{cycl}\frac{ab}{c}. \end{align}

Proof 5

As in Proof 3, the required inequality is reduced to

$\displaystyle\sum_{cycl}a^8b^8\ge a^4b^4c^4\sum_{cycl}a^2b^2.$

$\displaystyle 3\sum_{cycl}a^8b^8\ge \sum_{cycl}a^6b^6\sum_{cycl}a^2b^2.$

But, by the AM-GM inequality,

$\displaystyle \sum_{cycl}a^6b^6\ge 3\left(a^{12}b^{12}c^{12}\right)^{1/3}=3a^4b^4c^4.$

Acknowledgment

Dan Sitaru has kindly posted the above problem at the CutTheKnotMath facebook page, along with several solutions. The problem comes from his book Math Accent.

Proof 1 is by Ravi Prakash; Proof 2 is by Lâm Phan; Proof 3 is by Soumava Chakraborty; Proof 4 is by Rory Tarnow-Mordi.