# How prove this $\frac{1}{2a^2-6a+9}+\frac{1}{2b^2-6b+9}+\frac{1}{2c^2-6c+9}\le\frac{3} {5}\cdots (1)$

let $a,b,c$ are real numbers,and such $a+b+c=3$,show that $$\dfrac{1}{2a^2-6a+9}+\dfrac{1}{2b^2-6b+9}+\dfrac{1}{2c^2-6c+9}\le\dfrac{3} {5}\cdots (1)$$ I find sometimes,and I find this same problem:

let $a,b,c$ are real numbers,and such $a+b+c=3$,show that $$\frac{1}{5a^2-4a+11}+\frac{1}{5b^2-4b+11}+\frac{1}{5c^2-4c+11}\leq\frac{1}{4}$$ and this problem have some methods,you can see:http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=223910&start=20

and I like this can_hang2007 methods and Honey_S methods,But for $(1)$ I can't prove it.Thank you

• Are you allowed to use calculus? You can simply substitute a = 3 - b - c for all instances of a and then use calculus to proceed, however in the interest of time and solvability (i'm assuming have no calculators) that may not be practical. Sep 30, 2013 at 3:35
• I don't like calculus,because it must be very very ugly!! I like use AM-GM,cauchy-schwarz,jenson,and so on these nice methods solve inequality! Sep 30, 2013 at 3:39
• Are $a, b, c$ non-negative? Sep 30, 2013 at 4:04
• I think this equality is true for real numbers? Sep 30, 2013 at 4:11

Let $f(x) = \dfrac{1}{x^2+(3-x)^2}$. WLOG let $a \le b \le c$.

We note: $f(x) = f(3-x)$ and
if $x < 0$, then $f(x) < f(-x)$ as is obvious from signs or from $f(x)-f(-x) = \dfrac{12x}{4x^4+81}$. Using these, if $a < 0$, we also note $$f(a)+f(b)+f(c) < f(-a) + f(3-b)+f(3-c)$$ where the new arguments also fulfil the constraint.

Thus if $a < 0$, to prove the inequality holds for $(a, b, c)$, it is sufficient to show it holds for $(-a, 3-b, 3-c)$. If this also has negative variables, then one more application turns $(-a, 3-b, 3-c) \to (3+a, b, c-3)$, where the lower most value has increased by $3$ and the largest has decreased by $3$.

Using successive application of this as necessary, it is sufficient to consider cases where $a \ge 0$, i.e. where all the variables are non-negative. Now we look at the following version of the inequality,

$$-\sum_{cyc} \left(\frac{5}{2a^2-6a+9} - 1 \right) = \sum_{cyc} \frac{2(a-1)(a-2)}{2a^2-6a+9}$$

$$\frac{2(a-1)(a-2)}{2a^2-6a+9} = -\frac{2(a-1)}{5} + \frac{2(a-1)^2(2a+1)}{5(2a^2-6a+9)} \ge -\frac{2(a-1)}{5}$$

$$\implies -\sum_{cyc} \left(\frac{5}{2a^2-6a+9} - 1 \right) \ge \sum_{cyc}-\frac{2(a-1)}{5}=0$$

Hence proved.

• Is the assertion "if $x < 0$, then $f(-x) < f(x)$" always true? Oct 23, 2013 at 2:15
• @JosephQuinsey It is the other way round i.e. if $x < 0, f(x) < f(-x)$, which is used in the proof, I shall edit it in. Thanks. Oct 23, 2013 at 3:36
• Is a minus sign missing? $\sum_{cyc}\left(\frac{5}{2a^2-6a+9}-1\right)=\sum_{cyc}-\frac{2(a-1)(a-2)}{2a^2-6a+9}$. Oct 23, 2013 at 5:06
• $\frac{2(a-1)(a-2)}{2a^2-6a+9} = -\frac{2(a-1)}{5} + \frac{(a-1)^2(2a+1)}{5(2a^2-6a+9)}$ seems to be wrong. Oct 23, 2013 at 7:54
• @JosephQuinsey Check, the missing $2$ is edited. Oct 23, 2013 at 8:30

Let $f(x) = \dfrac{1}{2x^2-6x+9}$. First let us first quickly sketch some properties of $f$ that we'll need. $f(x)=\dfrac{1}{2(x-\frac32)^2+\frac92}$, so its maximum is $f(\frac32)=\frac29$. $f~'(x)=\dfrac{6-4x}{(2x^2-6x+9)^2}$, and $f~''(x)=12\dfrac{2x^2-6x+3}{(2x^2-6x+9)^3}$. Setting $f~''(x)=0$ gives the inflection points $x=\frac32\pm\frac{\sqrt3}2$. $f$ is increasing before $\frac32$, decreasing after $\frac32$, and is concave between the inflection points. Note $f(\frac32\pm\frac{\sqrt3}2)=\frac16$.

If $a$, $b$, and $c$ are all between the inflection points, then by Jensen's inequality $$f(a)+f(b)+f(c) \le 3f(\dfrac{a+b+c}3) = 3f(1) = \frac35$$ as required.

If at least two of $a$, $b$, or $c$ are outside the inflection points, then $$f(a)+f(b)+f(c) \le \frac16+\frac16+\frac29=\frac59\lt\frac35$$

If one of $a$, $b$, or $c$ is $\gt\frac32+\frac{\sqrt3}2$, then at least one of $a$, $b$, or $c$ is $\lt\frac32-\frac{\sqrt3}2$, otherwise $$3=a+b+c\ge 2(\frac32-\frac{\sqrt3}2)+(\frac32+\frac{\sqrt3}2)=\frac92-\frac{\sqrt3}2 \approx 3.7,$$ a contradiction. And so by the previous paragraph, the required result holds.

So the only remaining case is when exactly one of $a$, $b$, or $c$ is $\lt\frac32-\frac{\sqrt3}2$, and the other two are between the inflection points. We need two estimates. First consider the case when the minimum is $\le\frac12$. (Here $\frac12$ is arbitrary, although we need to choose something very close to $\frac12$ for the following argument to work.) Then $$f(a)+f(b)+f(c) \le f(\frac12) + \frac29 + \frac29 = \frac{2}{13} + \frac49 = \frac{70}{117} \lt \frac35$$

Finally, suppose $a$ is between $\frac12$ and $\frac32-\frac{\sqrt3}2$, and $b$ and $c$ are between the inflection points. By Jensen's inequality, we need only consider the case when $b=c$, by replacing them with their mean. Then $b=(3-a)/2$. So $a\gt\frac12$ implies $b\lt\frac54$, and $$f(a)+f(b)+f(b) \lt f(\frac32-\frac{\sqrt3}2) + 2f(\frac54) = \frac16 + \frac{16}{37} = \frac{133}{222} \lt \frac35$$

Here's what I've got so far. It's not a complete solution, but the ideas might be useful to someone else.

$2x^2-6x+9 =x^2+(x-3)^2$.

If $f(x) =2x^2-6x+9$, $f'(x) =4x-6$ is zero at $x = 3/2$. Since $f(3/2) =9/2$, $f(x) \ge 9/2$ for all $x$.

This means that $\dfrac1{f(x)} \le \dfrac{2}{9}$ for all $x$, so $\dfrac1{f(a)}+\dfrac1{f(b)}+\dfrac1{f(c)} \le \dfrac{6}{9} = \dfrac23$ without any hypothesis on $a, b, c$.

If $c = 3-a-b$, $f(c) =(3-a-b)^2+(a+b)^2$.

If $a=b=3/2$, $c = 0$ so $f(c) = 9$ so $\dfrac1{f(a)}+\dfrac1{f(b)}+\dfrac1{f(c)} = \dfrac{5}{9} < \dfrac{3}{5}$.

If $a=b=c$, $a=b=c=1$. Since $f(1) =1+4=5$, $\dfrac1{f(a)}+\dfrac1{f(b)}+\dfrac1{f(c)} = \dfrac{3}{5}$, so this is the equality case.

If we can show that $\dfrac1{f(a)}+\dfrac1{f(b)}+\dfrac1{f(3-a-b)}$ is a maximum when $a=b=1$, that would do it.

But I don't know how right now, so I'll leave it at this.

A comment to complete the other proof.

We try to show that it is enough to consider the inequality for only positive $a,b,c$. Assume $a\leq b\leq c$.

$$\sum_{cyc}\dfrac{1}{2a^2-6a+9}= \sum_{cyc}\dfrac{1}{(b+c)^2+a^2}= \sum_{cyc}\dfrac{1}{9-2ab-2ac}$$ Now it can be seen that if $a\leq 0$ and $b\leq 0$, one can choose $(-a,-b,c)$ instead and decrease the denominator and hence increase the whole sum (similar for the case where $a<0$).

In other words, for each triple $(a,b,c)$ for which at least one of $a,b,c$ is negative, one can find another triple with whole positive elements such that the sum is increased.

We need to prove that $$\sum_{cyc}\frac{1}{2a^2-6a+9}\leq\frac{3}{5}$$ or $$\sum_{cyc}\left(\frac{1}{5}-\frac{1}{2a^2-6a+9}\right)\geq0$$ or $$\sum_{cyc}\frac{a^2-3a+2}{2a^2-6a+9}\geq0$$ or $$\sum_{cyc}\left(\frac{18(a^2-3a+2)}{2a^2-6a+9}+1\right)\geq3$$ or $$\sum_{cyc}\frac{(2a-3)^2}{2a^2-6a+9}\geq\frac{3}{5}.$$ Now, by C-S we obtain: $$\sum_{cyc}\frac{(2a-3)^2}{2a^2-6a+9}=\sum_{cyc}\frac{(2a-3)^2(a-8)^2}{(2a^2-6a+9)(a-8)^2}\geq$$ $$\geq\frac{\left(\sum\limits_{cyc}(2a-3)(a-8)\right)^2}{\sum\limits_{cyc}(2a-3)^2(a-8)^2}=\frac{\left(\sum\limits_{cyc}(2a^2+5)\right)^2}{\sum\limits_{cyc}(2a-3)^2(a-8)^2}.$$ Hence, it remains to prove that $$\frac{\left(\sum\limits_{cyc}(2a^2+5)\right)^2}{\sum\limits_{cyc}(2a-3)^2(a-8)^2}\geq\frac{3}{5}.$$ Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative, and $abc=w^3$.

Hence, we need to prove a fourth degree polynomial inequality, id est, a linear inequality of $w^3$,

which says that it's enough to prove the last inequality for an extremal value of $w^3$,

which happens for equality case of two variables.

Let $b=a$ and $c=3-2a$.

We need to prove that $(a-1)^2(34a^2-98a+73)\geq0$, which is obvious.

Done!