$\displaystyle \frac{(1+a)(1+b)(1+c)}{(1-a)(1-b)(1-c)} < \frac{(2+a+b+c)^2}{(2-(a+b+c))^2}$
Solution
Lemma:
If $0 \leq p < q \leq r < s < 1$ such that $p+s = q+r$, then $\frac{(1+p)(1+s)}{(1-p)(1-s)} > \frac{(1+q)(1+r)}{(1-q)(1-r)}$
Proof:
$\displaystyle \frac{(1+p)(1+s)}{(1-p)(1-s)} > \frac{(1+q)(1+r)}{(1-q)(1-r)}$
$\displaystyle \iff \frac{1+p+s+ps}{1-(p+s)+ps} > \frac{1+q+r+qr}{1-(q+r)+qr}$
$\displaystyle \iff 1 + \frac{2(p+s)}{1-(p+s)+ps} > 1 + \frac{2(q+r)}{1-(q+r)+qr}$
$\displaystyle \iff \frac{2(p+s)}{1-(p+s)+ps} > \frac{2(q+r)}{1-(q+r)+qr}$
because, $p+s = q+r$,
$\displaystyle \iff \frac{1}{1-(p+s)+ps} > \frac{1}{1-(q+r)+qr}$
$\displaystyle \iff 1-(p+s)+ps < 1-(q+r)+qr$
$\iff ps < qr$, and substitute $p = q+r-s$,
$\iff (q+r-s)s < qr \iff (s-r)(s-q) > 0$ which is true.
Now to prove the original inequality, we use the usual substitution $a = x+y, b= y+z, c= x+z$ where $x,y,z > 0$. Since the perimeter $2(x+y+z)$ is less than 2, so we know that $x+y+z < 1$. So we know that $0 < x,y,z < 1$.
Our inequality now becomes:
$\displaystyle \frac{(1+x+y)(1+y+z)(1+x+z)}{(1-(x+y))(1-(y+z))(1-(x+z))} < \frac{(1+x+y+z)^2}{(1-(x+y+z))^2}$
Now we use the lemma:
$\displaystyle \frac{(1+x+y+z)(1+y)}{(1-(x+y+z))(1-y)} > \frac{(1+x+y)(1+y+z)}{(1-(x+y))(1-(y+z))} $
and again using the lemma:
$\displaystyle \frac{(1+x+y+z)(1+0)}{(1-(x+y+z))(1-0)} > \frac{(1+y)(1+x+z)}{(1-y)(1-(x+z))} $
Multiplying the last two inequalities gives us the desired result.
Remark: A lot of people commented that the lemma is too "out of the blue" or "out of the magician's hat." The key thing to notice here is the form $\frac{(1+x)(1+y)}{(1-x)(1-y)}$ is minimized when $x$ and $y$ are closer together, and that observation is at the heart of the lemma. On the original inequality's LHS, we have the product of $a,b,c$ in that form, while on the RHS we have $(a+b+c)/2$, $(a+b+c)/2$, 0. Which triple is closer together and which is farther apart?
No comments:
Post a Comment