An Inequality in Three (Or Is It Two) Variables

Problem

An Inequality in Three (Or Is It Two) Variables

Solution 1

Denote $\sin^2 z=a; \cos^2 z=b.\,$ Then

$\displaystyle \begin{align} &\frac{x}{ay+bz}+\frac{y}{az+bx}+\frac{z}{ax+by}\\ &\qquad\qquad\qquad=\frac{x^2}{axy+bxz}+\frac{y^2}{ayz+bxy}+\frac{z^2}{axz+byz}\\ &\qquad\qquad\qquad\geq \frac{(x+y+z)^2}{(xy+yz+zx)(a+b)}\\ &\qquad\qquad\qquad\geq \frac{3(xy+xz+yz)}{(xy+yz+zx)(a+b)}=\frac{3}{a+b} \end{align}$

so that

(1)

$\displaystyle \frac{x}{ay+bz}+\frac{y}{az+bx}+\frac{z}{ax+by}\ge\frac{3}{a+b}.$

For $z=x\,$ in (1):

$\displaystyle \frac{x(ax+by)+x(ay+bx)}{(ay+bx)(ax+by)}+\frac{y}{x(a+b)}\geq \frac{3}{a+b}\\ \displaystyle \frac{x(ax+ay+bx+by)}{(ay+bx)(ax+by)}+\frac{y}{x(a+b)}\geq \frac{3}{a+b}.$

(2)

$\displaystyle \frac{x(a+b)(x+y)}{(ax+by)(bx+ay)}+\frac{1}{a+b}\cdot \frac{y}{x}\geq \frac{3}{a+b}.$

Similarly, with $z=y\,$ in (1):

(3)

$\displaystyle \frac{y(a+b)(x+y)}{(ax+by)(bx+ay)}+\frac{1}{a+b}\cdot \frac{x}{y}\geq \frac{3}{a+b}.$

By adding (2) and (3):

(4)

$\displaystyle \frac{(a+b)(x+y)^2}{(ax+by)(bx+ay)}+\frac{1}{a+b}\Bigr(\frac{x}{y}+\frac{y}{x}\Bigr)\geq \frac{6}{a+b}.$

Now, replace back $a=\sin^2 z;\,b=\cos^2 z\,$ in (4):

$\displaystyle \frac{(\sin^2 z+\cos^2 z)(x+y)^2}{(x\sin^2 z+y\cos^2 z)(x\cos^2 z+y\sin^2 z)}+\frac{x}{y}+\frac{y}{x}\geq 6.$

Equality holds for $x=y.$

Solution 2

Let's prove a little more general result:

If $x,y,a,b\gt 0\,$ and $a+b=1:\,$

$\displaystyle \frac{(x+y)^2}{(xa+yb)(xb+ya)}+\frac{x}{y}+\frac{y}{x}\geq 6.$

By the AM-GM inequality $\displaystyle ab\le \frac{1}{4}\,$ and $\displaystyle a^2+b^2\le \frac{1}{2}(a+b)^2=\frac{1}{2}.\,$ Using that

$\displaystyle \begin{align} (xa+yb)(xb+ya)&= x^2ab+xyb^2+xya^2+y^2ab\\ &\le \frac{1}{4}(x^2+y^2)+\frac{1}{2}xy=\frac{1}{4}(x+y)^2. \end{align}$

Thus

$\displaystyle\begin{align} &\frac{(x+y)^2}{(x\sin^2z+y\cos^2z)(x\cos^2z+y\sin^2z)}+\frac{x}{y}+\frac{y}{x}\\ &\qquad\qquad\qquad\ge\frac{\displaystyle (x+y)^2}{\displaystyle \frac{(x+y)^2}{4}}+\frac{x}{y}+\frac{y}{x}\\ &\qquad\qquad\qquad=4+2=6. \end{align}$

Solution 3

Let $a=x\sin^2z+y\cos^2z,\,$ $b=y\sin^2z+x\cos^2z.\,$ Then $x+y=a+b\,$ and the inequality to prove becomes

$\displaystyle \frac{(a+b)^2}{ab}+\frac{x}{y}+\frac{y}{x}\ge 6.$

This is the same as

$\displaystyle \frac{a}{b}+\frac{b}{a}+2+\frac{x}{y}+\frac{y}{x}\ge 2+2+2=6.$

Acknowledgment

This problem from the Romanian Mathematical Magazine has been kindly posted at the CutTheKnotMath facebook page by Dan Sitaru. He later mailed his solution (Solution 1) in a LaTeX file. Solution 3 is by Ravi Prakash and independently by Kevin Soto Palacios.

 

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