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I've spent some time on this problem but I can't find a method to solve it:

Let $x, y, z>0$, prove that: $$\frac{xy}{xy+x+y} + \frac{yz}{yz+y+z} + \frac{zx}{zx+z+x} \le \frac{6+x^2+y^2+z^2}{9}$$

That is how I thought :

first of all, if we have $a, b, x>0$ and $a\le b$, then $$\frac{a}{a+x}\le\frac{b}{b+x}$$ (which can be simply proved by making the product on diagonal)

So I try to find a $b$ such as $b\ge xy$.

From means inequality (geometric mean and root mean square), we have $$\sqrt{xy} \le \sqrt{\frac{x^2+y^2}{2}}$$ or $$xy \le \frac{x^2+y^2}{2}$$ So $$\frac{xy}{xy+x+y} \le \frac{\frac{x^2+y^2}{2}}{\frac{x^2+y^2}{2}+x+y}$$ Now, I need to find something which is $\le x+y$ and that's where I've got stuck.

Some hints would be really apreciated! Thanks!

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    from where do you got that? – 2017-01-07
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    @Dr.SonnhardGraubner the problem? I've found it in an old book for preparing for math olympiad/contests – 2017-01-07
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    and the name of the book is? – 2017-01-07
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    Maybe you can adjust the solution of this inequality to your needs: $$\frac{3xy}{xy+x+y}+ \frac{3yz}{yz+y+z}+ \frac{3zx}{zx+z+x}\leq 2+ \frac{x^2+y^2+z^2}{3}$$ Solution: \begin{align*} 2+ \frac{x^2+y^2+z^2}{3} &=\sum \frac{x^2+y^2+4}{6} \\ &\geq \sum \sqrt[6]{x^2y^2} \\ &=\sum \frac{xy}{\sqrt[3]{(xy)xy}} \\ &\geq \sum \frac{xy}{\left ( xy+x+y \right )/3}\\ &=\sum \frac{3xy}{xy+x+y} \end{align*} – 2017-01-07
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    @Dr.SonnhardGraubner "Algebraic inequalities - from inititiation to performance" (it isn't an English book, I've translated the title) – 2017-01-07
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    @Tolaso thanks a lot! it worked – 2017-01-07
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    @Tolaso Perhaps you could post that as an answer. Just divide through by $3$. – 2017-01-07
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    @MichaelBurr Sure! – 2017-01-07

2 Answers 2

4

Equivenantly we have to prove that

$$\frac{3xy}{xy+x+y}+ \frac{3yz}{yz+y+z}+ \frac{3zx}{zx+z+x}\leq 2+ \frac{x^2+y^2+z^2}{3}$$

Solution

\begin{align*} 2+ \frac{x^2+y^2+z^2}{3} &=\sum \frac{x^2+y^2+4}{6} \\ &\geq \sum \sqrt[6]{x^2y^2} \\ &=\sum \frac{xy}{\sqrt[3]{(xy)xy}} \\ &\geq \sum \frac{xy}{\left ( xy+x+y \right )/3}\\ &=\sum \frac{3xy}{xy+x+y} \end{align*}

It only makes use of AM-GM inequality.. :)

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Here is an other elementary one.

Since $\displaystyle{\frac{1}{x}+\frac{1}{y}\geq\frac{4}{x+y}}$, we have that:

$\displaystyle{\frac{xy}{xy+x+y}=\frac{1}{1+\frac{1}{x}+\frac{1}{y}}\leq \frac{1}{1+\frac{4}{x+y}}=1-\frac{4}{4+x+y}}$

Similarly we get the other two inequalities. Thus it suffices to prove that:

$\displaystyle{\frac{6+x^2+y^2+z^2}{9}\geq 1-\frac{4}{4+x+y}+1-\frac{4}{4+y+z}+1-\frac{4}{4+z+x}\Leftrightarrow \\ 5+\frac{x^2+y^2}{2}+5+\frac{y^2+z^2}{2}+5+\frac{z^2+x^2}{2}+\frac{36}{x+y+4}+\frac{36}{y+z+4}+\frac{36}{z+x+4}\geq 36}$

But

$\displaystyle{\frac{x^2+y^2}{2}+1=\frac{x^2+1}{2}+\frac{y^2+1}{2}\geq x+y}$ and so:

$\displaystyle{5+\frac{x^2+y^2}{2}+\frac{36}{x+y+4}\geq (5+\frac{x^2+y^2}{2})+\frac{36}{(5+\frac{x^2+y^2}{2})}\geq2\sqrt{(5+\frac{x^2+y^2}{2})\frac{36}{(5+\frac{x^2+y^2}{2})}}=12}$

Simlarly we get the other two inequalities with $y,z$ and $z,x$, we add them and we are done.

Note: No need of AM-GM for $6$ terms.