# An Inequality in Reciprocals II

### Problem

### Solution 1

Set $x=4a+1\ge 1,\,$ $y=4b+1\ge 1,\,$ $x+y=2(2a+2b+1).\,$ Further,

$\begin{align} 12a+4b+4 &= 3(4a+1)+(4b+1)=3x+y,\\ 12b+4a+4 &= 3(4b+1)+(4a+1)=3y+x.\\ \end{align}$

We have to prove that

$\displaystyle\frac{1}{x}+\frac{1}{y}+\frac{12}{x+y}\ge\frac{16}{3x+y}+\frac{16}{3y+x}.$

This is the same as $\displaystyle\frac{(x+y)^2+12xy}{xy(x+y)}\ge\frac{64(x+y)}{(3x+y)(3y+x)},\,$ or

$\displaystyle\frac{(x+y)^2+12xy}{xy(x+y)}\ge\frac{64(x+y)}{3x^2+3y^2+10xy}.$

Multiplying out, expanding, and eliminating like terms, we reduce the latter to $(x-y)^4\ge 0,\,$ which is obviously true. Equality is achieved for $x=y,\,$ i.e., $a=b.$

### Solution 2

The required inequality is equivalent to an inequality in integrals:

$\displaystyle\begin{align}\int_{0}^{1}x^{4a}dx+\int_{0}^{1}x^{4b}dx&+6\int_{0}^{1}x^{2(a+b)}dx\\ &\ge 4\int_{0}^{1}x^{3a+b}dx+4\int_{0}^{1}x^{3b+a}dx, \end{align}$

which can be rewritten (now without the integrals) as

$A^4+b^4+6A^2B^2\ge4AB(A^2+B^2),$

or, equivalently,

$(A^2+B^2)^2-4AB(A^2+B^2)+4A^2B^2\ge 0,$

i.e., $(A-B)^4\ge 0,\,$ which is true.

### Solution 3

The required inequality is equivalent to

$\displaystyle\int_{0}^{1}(t^{4a}+t^{4b}+6t^{2(a+b)})dt\ge 4\int_{0}^{1}(t^{3a+b}+t^{3b+a})dt.$

By letting $x=t^a\,$ and $y=t^b\,$ this reduces to $x^4+y^4+6x^2y^2\ge 4xy(x^2+y^2),\,$ or $(x-y)^4\ge 0.$

### Solution 4

Let $A=\displaystyle\frac{1}{4a+1} + \frac{3}{2a+2b+1}-\frac{4}{3a+b+1},\,$ $B=\displaystyle\frac{1}{4b+1} + \frac{3}{2a+2b+1}-\frac{4}{3b+a+1}.\,$ We find

$\displaystyle\begin{align} A&=\frac{2(a-b)(5a-b-1)}{(4a+1)(3a+b+1)(2a+2b+1)},\\ B&=\frac{2(a-b)(a-5b-1)}{(4b+1)(3b+a+1)(2a+2b+1)} \end{align}$

such that

$\displaystyle\begin{align} A+B &= \frac{2(a-b)}{2a+2b+1}\left(\frac{5a-b-1}{(4a+1)(3a+b+1)}+\frac{a-5b-1}{(4b+1)(3b+a+1)}\right)\\ &=\frac{24(a-b)^{4}}{D}, \end{align}$

where $D=(4a+1)(4b+1)(3a+b+1)(a+3b+1)(2a+2b+1).\;$ Clearly $A+B\ge 0.\,$ Equality is achieved when $a=b.$

### Illustration

LHS in blue, RHS in red.

### Acknowledgment

Dan Sitaru has kindly communicated to me the above problem, along with three solution. This is problem JP032 from the Romanian Mathematical Magazine, where it accumulated four solutions. I reproduce these here. The inequality is an invention of Hung Nguyen Viet (Vietnam). Solution 1 is by Kevin Soto Palacios (Peru); Solution 2 is by Soumitra Mandal (India); Solution 3 is by Henry Ricardo (USA); Solution 4 is by Imad Zak (Lebanon). The illustration is by Gary Davis.

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