6

let $a,b,c>0$ ,and such $a+b+c=3$.prove that $$\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{1+c^2}<\dfrac{11}{5}$$

if this problem find this minimum,then $$f(x)=\dfrac{1}{x^2+1}\ge ax+b$$ where $a=f'(1),a+b=f(1)$

since $$f'(x)=-\dfrac{2x}{(1+x^2)^2}\Longrightarrow f'(1)=-\dfrac{1}{2}$$ so $$a=-\dfrac{1}{2},b=1$$

and $$\dfrac{1}{1+x^2}+\dfrac{1}{2}x-1=\dfrac{x(x-1)^2}{(x^2+1)^2}\ge 0$$ then $$\dfrac{1}{1+x^2}\ge-\dfrac{1}{2}x+1$$ so $$\sum_{cyc}\dfrac{1}{1+a^2}\ge\sum_{cyc}-\dfrac{1}{2}a+1=-\dfrac{1}{2}(a+b+c)+3=\dfrac{3}{2}$$

But for maximum,I can't prove it.Thank you

math110
  • 93,304
  • If this (http://math.stackexchange.com/questions/545810/extreme-value-of-increasing-or-decreasing-function) is true, the maximum value will given by $a=b=c$ – lab bhattacharjee Nov 14 '13 at 04:00
  • No,I think this problem maximum not $a=b=c$.But Thank you all the same – math110 Nov 14 '13 at 04:02
  • When in doubt, just plot it out. The maximum $\frac{21}{10}$ is achieved when two of $a,b,c$ is $0$ while the remaining one is $3$. Whether there is an elegant proof for this or not. I've no idea. – achille hui Nov 14 '13 at 04:19
  • @achillehui: taking the limit as $(a,b)\to (0,0)$ should suffice to show that value as a possible solution... – abiessu Nov 14 '13 at 04:22
  • 2
    @achillehui There are some pitfalls in plotting though. I get a value higher than $2.1$ using $a=2.94, b = 0.03, c = 0.03$, which doesn't appear to be round off error. So the answer may be in the neighbourhood of $(3, 0, 0)$, though not exactly so. – Macavity Nov 14 '13 at 04:24
  • 1
    @Macavity, you are right. The maxmimum seems to lie at points like $(x,x,3-x)$ where $x \sim 0.031776261136413$, a root of the equation $6 x^4-27 x^3+49 x^2-33 x+1$. – achille hui Nov 14 '13 at 04:41
  • 1
    @achillehui Wow. I think elegant proof just flew out of the window. Looks like Lagrange to the rescue. – Macavity Nov 14 '13 at 04:43
  • Now, I have edit this problem,Now we can solve it? – math110 Nov 14 '13 at 04:54
  • $f(x) = \frac{1}{x^2+1}$ has only one inflection point in $[0,3]$. By playing with convexity it's easy to deduce that two of the variables are equal, then reduce it to a one-variable inequality. –  Nov 14 '13 at 05:00
  • Hello,Now this inequality is true? – math110 Nov 14 '13 at 05:02
  • 1
    LHS is $\approx 2.1019 > 2.1$ when $a \approx 2.93645, b \approx 0.0317763, c \approx 0.0317763$. So the the bound is not correct. – Macavity Nov 14 '13 at 05:04
  • But,Now $\le\dfrac{11}{5}$ is true – math110 Nov 14 '13 at 05:09
  • Yes - and that's the same as showing $\frac{11}{15} - \frac1{x^2 + 1} - \frac{10}{33} (x - 1)$ is positive, which is quite easy compared to the original problem. – Macavity Nov 14 '13 at 05:21
  • the bound is 2.1019 as macavity said,2.2 is much easier. – chenbai Nov 14 '13 at 06:52

2 Answers2

5

First show that $\dfrac{1}{1+x^2} < \dfrac{63 - 19x}{60}$ for $0 < x < 3$.

Then $\text{LHS} < \dfrac{3\cdot63 - 19(a + b + c)}{60} = \dfrac{189 - 57}{60} = \text{RHS}$.

Christopher Toni
  • 2,025
apt1002
  • 2,033
4

By symmetry, it just suffices to look at the case $0 \le a \le b \le c$ where $a + b + c = 3$.

Let $f(x) = \frac{1}{1+x^2}$ and $F(a,b,c) = f(a)+f(b)+f(c)$. Notice

$$\frac{d^2}{dx^2} f(x) = \frac{2(3x^2-1)}{(1+x^2)^3} \longrightarrow \begin{cases} > 0,&\text{ for } x > \frac{1}{\sqrt{3}}\\< 0,&\text{ for } x <\frac{1}{\sqrt{3}}\end{cases}$$ The function $f(x)$ is concave on $[0,\frac{1}{\sqrt{3}}]$ and convex on $[\frac{1}{\sqrt{3}},3]$.

Since $a + b + c = 3$, we know $c \ge 1$. This implies $c \in (\frac{1}{\sqrt{3}}, 3]$ where $f(x)$ is convex. If $b$ also belongs to this interval, then we can increase the value of $F(a,b,c)$ by decreasing $b$ and increasing $c$ symmetrically w.r.t their mean $\frac{3-a}{2}$ until either $b$ hits $\frac{1}{\sqrt{3}}$ or $c$ hits $3$. It is easy to see $b$ will hit $\frac{1}{\sqrt{3}}$ first. From this, we can conclude when $b > \frac{1}{\sqrt{3}}$,

$$F(a,b,c) \le F(a, \frac{1}{\sqrt{3}}, 3-a-\frac{1}{\sqrt{3}})$$

If $a > \frac{1}{\sqrt{3}}$, the new $b$ will be smaller than $a$. We can flip the roles of $a$ and $b$ and repeat above trick. At the end, we will find a pair of $a', b' \in [0, \frac{1}{\sqrt{3}}]$ such that

$$F(a,b,c) \le F(a',b',c')\quad\text{ where }\;\;c' = 3 - a'-b'$$

Since $f(x)$ is concave there, we have

$$F(a',b',c') \le F(\frac{a'+b'}{2},\frac{a'+b'}{2},c')$$

This means in order to find the optimal $(a,b,c)$ which maximize $F(a,b,c)$, one only need to look at those points of the form

$$(a,b,c) = (x,x,3-2x)\quad\text{ where } x \in [0,\frac{1}{\sqrt{3}}]$$

For such points, $F(a,b,c)$ reduces to

$$F(x,x,3-2x) = \frac{2}{1+x^2} + \frac{1}{1 + (3-2x)^2}$$ Let us call this function as $G(x)$. Notice

$$\frac{dG(x)}{dx} = \frac{4(3-2x)}{(1+(3-2x)^2)^2} - \frac{4x}{(1+x^2)^2}$$ We have $$G'(0) = \frac{3}{25}> 0\quad\text{ and }\quad G'(\frac{1}{\sqrt{3}}) = \frac{1944\sqrt{3}-3645}{1588\sqrt{3}-2448} \sim -0.91866538126469 < 0$$ So the optimal $(a,b,c) = (x,x,3-2x) $ occurs for a $x \in (0,\frac{1}{\sqrt{3}})$. We can determine the corresponding value of $x$ by solving

$$G'(x) = 0\quad\iff\quad 6x^4-27x^3+49x^2-33x+1 = 0$$

The polynomial in R.H.S has 2 real roots, only one of it $\sim 0.031776261136413$ falls inside the interval $[0,\frac{1}{\sqrt{3}}]$. So the maximum value of $F(a,b,c)$ is about $$\max F(a,b,c) \sim G(0.031776261136413) \sim 2.101903255548146 < \frac{11}{5}$$

achille hui
  • 122,701