Three Junior Problems from Vietnam

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Three Junior Problems from Vietnam

Preliminaries

First we'll express $a,b,c\;$ in terms of $x,y,z.\;$ The relationship can be expressed in matrix terms as

$\left(\begin{array}{ccc}0&y&z\\x&0&z\\x&y&0\end{array}\right)\left(\begin{array}\;a\\b\\c\end{array}\right)=\left(\begin{array}\;x\\y\\z\end{array}\right).$

In the standard way,

$\left(\begin{array}{ccc}0&y&z\\x&0&z\\x&y&0\end{array}\right)^{-1}=\displaystyle\frac{1}{2xyz}\left(\begin{array}{ccc}-yz&yz&yz\\xz&-xz&xz\\xy&xy&-xy\end{array}\right),$

so that

$\displaystyle\begin{align} a&=\frac{-x+y+z}{2x}\\ b&=\frac{x-y+z}{2y}\\ c&=\frac{x+y-z}{2z}. \end{align}$

Now, we may observe that if a triple $\{x,y,z\}\;$ satisfies the three conditions for the given $a,b,c,\;$ then so does the triple $\{-x,-y,-z\}.\;$ I claim that the three parameters are bound to be of the same sign (so that, if need be, we may choose all three to be positive.) Indeed, if say, $x\lt 0\;$ but $y,z\gt 0\;$ then $a=\displaystyle\frac{-x+y+z}{2x}\lt 0\;$ which contradicts that $a\;$ is given to be positive. Similarly, if say, $x,y\lt 0\;$ and $z\gt 0,\;$ then $c=\displaystyle\frac{x+y-z}{2z}\lt 0\;$ which is also impossible.

Thus, in the proofs I shall assume that $x,y,z\gt 0.\;$ I thank Professor Sam Walters (UNBC) for pointing out the need to discuss the signs of those parameters.

Proof of a.

The inequality $\displaystyle abc\le\frac{1}{8}\;$ is equivalent to

$(-x+y+z)(x-y+z)(x+y+z)\le xyz.$

Let's set $u=-x+y+z,\;$ $v=x-y+z,\;$ $w=x+y-z.\;$ Then we have $x=\displaystyle\frac{v+w}{2},\;$ $y=\displaystyle\frac{w+u}{2},\;$ $z=\displaystyle\frac{u+v}{2}.\;$ Note that, since $a,b,c\;$ and $x,y,z\;$ are positive, so are $u,v,w.\;$ The required inequality becomes

$\displaystyle uvw\le\frac{u+v}{2}\frac{v+w}{2}\frac{w+u}{2}.\;$

This is true because $\displaystyle\sqrt{uv}\le\frac{u+v}{2},\;$ $\displaystyle\sqrt{vw}\le\frac{v+w}{2},\;$ $\displaystyle\sqrt{wu}\le\frac{w+u}{2},$ which, when multiplied, give the above inequality.

Proof of b.

In the notations as above,

$\displaystyle\begin{align}2+a+b &= 2+\frac{1}{2}\left(-1+\frac{y}{x}+\frac{z}{x}\right)+\frac{1}{2}\left(\frac{x}{y}-1+\frac{z}{y}\right)\\ &=1+\frac{1}{2}\left(\frac{y}{x}+\frac{z}{x}\right)+\frac{1}{2}\left(\frac{x}{y}+\frac{z}{y}\right)\\ &=\frac{2xy+y^2+yz+x^2+xz}{2xy}\\ &=\frac{(x+y+z)(x+y)}{2xy},\\ 2+b+c&=\frac{(x+y+z)(y+z)}{2yz},\\ 2+c+a&=\frac{(x+y+z)(z+x)}{2xz}. \end{align}$

It follows that

$\displaystyle\begin{align} \frac{1}{2+a+b}+\frac{1}{2+b+c}+\frac{1}{2+c+a} &= \frac{2}{x+y+z}\sum_{cycl}\frac{xy}{x+y}\\ &\le\frac{2}{x+y+z}\sum_{cycl}\frac{\displaystyle\left(\frac{x+y}{2}\right)^2}{x+y}\\ &=\frac{2}{x+y+z}\sum_{cycl}\frac{x+y}{4}\\ &=\frac{2}{x+y+z}\frac{(x+y)+(y+z)+(z+x)}{4}\\ &=1. \end{align}$

Proof of c.

As in proof of a, we set $u=-x+y+z,\;$ $v=x-y+z,\;$ $w=x+y-z\;$ and get $\,2x=v+w,\;$ $2y=w+u,\;$ $2z=u+v.\;$ The required inequality becomes

$\displaystyle \sum_{cycl}\frac{u}{v+w}\ge 2\sum_{cycl}\frac{u}{v+w}\frac{v}{w+u}.$

Getting rid of the denominators leads to the equivalent

$\displaystyle \sum_{cycl}u(u+v)(u+w)\ge 2\sum_{cycl}uv(u+v),$

or,

$\displaystyle \sum_{cycl}(u^3+uvw)\ge \sum_{cycl}uv(u+v).$

This is exactly Schur's inequality for $t=1.$

Acknowledgment

The problem is due to Nguen Viet Hung (Vietnam); it was included as JP.026 at the ssmrmh.ro and posted at the CutTheKnotMath facebook page by Dan Sitaru.

 

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