Can You Solve This Challenging Inequality Involving Real Numbers?

[anemone]

Here is this week's POTW:

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The reals $x$ and $y$ are such that $0 < x< 1$ , and $y> 0$, prove that

\(\displaystyle (x+ y)\left(\frac{1}{x}+\frac{1}{y} -\frac{4}{(x+1)^2}\right) ≥ \frac{4}{(x+1)^2}.\)

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

No one answered last week problem. :(

You can see my solution as follows:

Spoiler

Since $(x+1)^2>0$ and $xy>0$, we can multiply through the inequality by $xy(x+1)^2$, and it remains to prove $(x + y)^2(x+1)^2− 4xy(x+y) ≥ 4xy$, but note that

\(\displaystyle \begin{align*}(x + y)^2(x+1)^2− 4xy(x+y) -4xy&=(x^2+xy+x+y)^2-4xy(x+y+1)\\&=(x(x+y+1)+y)^2-4xy(x+y+1)\\&=x^2(x+y+1)^2+2xy(x+y+1)+y^2-4xy(x+y+1)\\&=x^2(x+y+1)^2-2xy(x+y+1)+y^2\\&=(x(x+y+1)-y)^2\end{align*}\)

and this quantity is definitely greater than or equals to zero, and the result follows.