An Inequality in Reciprocals II

Problem

An  Inequality in Reciprocals II

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.

An  Inequality in Reciprocals II, illustration

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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