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If $a+b+c+d = 2$, prove that

$$\dfrac{a^2}{(a^2+1)^2}+\dfrac{b^2}{(b^2+1)^2}+\dfrac{c^2}{(c^2+1)^2}+\dfrac{d^2}{(d^2+1)^2}\le \dfrac{16}{25}$$

Also $a,b,c,d \ge 0$.

Bart Michels
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todestrieb
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  • what did you try already? – kazekage Mar 29 '14 at 13:32
  • I haven't tried anything worth mentioning frankly. :) – todestrieb Mar 29 '14 at 13:45
  • Young's Inequality (http://en.wikipedia.org/wiki/Young%27s_inequality) might be somehow useful, although I have not figured out how. – Yiyuan Lee Mar 29 '14 at 15:00
  • If you can prove $$\frac{a^2}{(a^2+1)^2}+\frac{b^2}{(b^2+1)^2}\le2 \frac{\left(\frac{a+b}2\right)^2‌​}{\left(\left(\frac{a+b}2\right)^2+1\right)^2}$$ for $1\le a+b\le2$ which is the same as $\frac{4a^2}{(4a^2+1)^2}+\frac{4b^2}{(4b^2+1)^2}\le\frac{2(a+b)^2}{((a+b)^2+1)^2‌​}$ for $\frac12\le a+b\le1$, then WLOG $b=c=d$ by repeated usage and continuity argument. This inequality is true, but I cannot prove it nicely. – user2345215 Mar 30 '14 at 08:41

2 Answers2

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Assume $0 \le a \le b \le c \le d$, with $a+b+c+d=2$,

Then $(48a-4)(a^2+1)^2-125a^2 = (2a-1)^2(12a^3+11a^2+32a - 4) \ge 0$, for $a \ge \frac{1}{8}$;

(See that $32a-4 \ge 0$ and $12a^3+11a^2$ is positive).

That is $\dfrac{a^2}{(a^2+1)^2} \le \dfrac{48a-4}{125}$, and similarly for $b,c,d$ and adding them we have $\sum\limits_{a,b,c,d} \dfrac{a^2}{(a^2+1)^2} \le \sum\limits_{a,b,c,d} \dfrac{48a-4}{125}=\dfrac{80}{125}=\dfrac{16}{25}$.

The case $a < \frac{1}{8}$,

We have $(540x + 108)(x^2+1)^2 - 2197x^2 = (3x-2)^2(60x^3+92x^2+216x+27) \ge 0$, for $x \ge 0$.

Thus $\sum\limits_{x \in \{b,c,d\}} \dfrac{x^2}{(x^2+1)^2} \le \sum\limits_{x \in \{b,c,d\}} \dfrac{540x+108}{2197}=\dfrac{108}{169}-\dfrac{540a}{2197}$

and, $\dfrac{a^2}{(a^2+1)^2} < a^2 < \dfrac{a}{8} < \dfrac{540a}{2197}$.

So, $\sum\limits_{a,b,c,d} \dfrac{a^2}{(a^2+1)^2} \le \dfrac{108}{169} < \dfrac{16}{25}$

Equality occurs iff $a=b=c=d=\dfrac{1}{2}$.

r9m
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  • Why does $$\sum\limits_{x \in {b,c,d}} \dfrac{540x+108}{2197}=\dfrac{108}{169}-\dfrac{540a}{2197}?$$ Shouldn't it be $$\sum\limits_{x \in {b,c,d}} \dfrac{540x+108}{2197}=\dfrac{324}{2197}-\dfrac{540a}{2197} \lt \dfrac{16}{25}-\dfrac{540a}{2197}$$

    Also where did you get the insight for the equations you used?

     – todestrieb Mar 30 '14 at 06:45
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    @todestrieb It's correct $$\sum_{x\in{b,c,d}}\frac{540x+108}{2197}=\frac{540(b+c+d)+324}{2197}=\frac{54!!0(2-a)+324}{2197}=\frac{1404}{2197}-\frac{540a}{2197}=\frac{108}{169}-\frac{5!!40a}{2197}$$ As for the insight I think it's pretty clear he wanted to find linear estimates. That can take a little work, but with computers it's pretty easy to verify if you are correct. – user2345215 Mar 30 '14 at 09:03
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    You are both right. I am thinking of a generalization of this. Is the following true? $$\dfrac{x_1^2}{(x_1^2+1)^2}+\dfrac{x_2^2}{(x_2^2+1)^2}+\cdot\cdot\cdot+\dfrac{x_n^2}{(x_n^2+1)^2}\le \dfrac{n^2}{(n+1)^2}$$ with $$x_1+x_2+\cdot\cdot\cdot+x_n= \sqrt{n}$$$$x_1,x_2,\cdot\cdot\cdot,x_n \ge0$$$$ n \in \mathbb{N}$$ – todestrieb Mar 31 '14 at 07:04
  • If you wish to answer go here: http://math.stackexchange.com/questions/733675/inequality-with-x-1x-2-cdot-cdot-cdotx-n-sqrtn – todestrieb Mar 31 '14 at 10:07
1

Let us consider the function $$ \frac{x^2}{(1+x^2)^2} + \frac{y^2}{(1+y^2)^2} + \frac{z^2}{(1+z^2)^2} + \frac{t^2}{(1+t^2)^2} + \lambda(x+y+z+t-2) $$and write the first order conditions: $$ 0 = -\frac{2x(x^2-1)}{(1+x^2)^3} + \lambda, $$and the same equation for $y,z,t$. In particular, $$ \frac{x(x^2-1)}{(1+x^2)^3} = \frac{y(y^2-1)}{(1+y^2)^3} = \frac{z(z^2-1)}{(1+z^2)^3} = \frac{t(t^2-1)}{(1+t^2)^3} $$

Taking a look at the variations of this function,

we see that if $x\neq y$ then $x,y < 1$ and $x,y,z,t\in \{u,v\}$, with

$$ \begin{cases} \frac{u(u^2-1)}{(1+u^2)^3} = \frac{v(v^2-1)}{(1+v^2)^3} \\ u+v=1\\ u<v. \end{cases} $$ Then an analysis proves that $(u,v) = (0,1)$:

Eventually, compute the two potential extrema: $$ \frac{1^2}{(1+1^2)^2} + \frac{1^2}{(1+1^2)^2}+0+0 =\frac 12;\\ \frac{(1/2)^2}{(1+(1/2)^2)^2} + \frac{(1/2)^2}{(1+(1/2)^2)^2}+ \frac{(1/2)^2}{(1+(1/2)^2)^2} + \frac{(1/2)^2}{(1+(1/2)^2)^2} = \frac {16}{25}. $$

mookid
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