Maximising Can Also Be Difficult!

Algebra Level 3

$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} = a+b+c$

Let $a,b,c$ be positive real numbers such that the above equation is satisfied. If the maximum value of the expression below is in the form of $\frac {m}{n} ,$ where $m,n$ are coprime positive integers, what is $m+n?$

$\dfrac{1}{\left(2a+b+c\right)^2}+\dfrac{1}{\left(2b+c+a\right)^2}+\dfrac{1}{\left(2c+b+a\right)^2}$

The answer is 19.

4 solutions

Kartik Sharma
Mar 16, 2015

We know by Titu Cauchy(or whatever the name is, I don't remember the name, only the identity, or even harmonic) that

$\displaystyle (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})(a + b + c) \geq {3}^{2}$

Therefore, $\displaystyle (a + b +c) \geq 3$

$\displaystyle \sum_{cyc}{\frac{1}{{(2a + b +c)}^{2}}}$ can be written as $\displaystyle \sum_{cyc}{\frac{1}{{(S + a)}^{2}}}$ for $\displaystyle S = a + b + c = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$

Now Jensen's inequality can easily be used and yes you might argue that this is a convex function but remember taking double derivative for $\displaystyle S = a + b + c$ , it is but for $\displaystyle S = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ , it is concave, it will come out to be less than or equal to zero. (Even we can verify concavity using the definition)

Hence

$\displaystyle \sum_{cyc}{f(a)} \le 3f(\frac{a + b + c}{3})$ for $\displaystyle f(x) = \frac{1}{{(S + x)}^{2}}$

$\displaystyle \le 3\frac{1}{{(S + \frac{S}{3})}^{2}}$

$\displaystyle \le 3\frac{9}{16{S}^{2}}$

and we know $\displaystyle S \geq 3$

therefore, $\displaystyle \le 3\frac{9}{16*9} = \frac{3}{16}$

and of course there is equality at $\displaystyle (1,1,1)$ -_-

Cauchy Schwartz Inequality wasn't bad.. But this question just required AM greater equal to HM (equality holds when identical no's.)😊 AM Arithmetic Mean HM Harmonic

Amartya Anshuman - 5 years, 10 months ago

Can someone explain me why the function is concave?

Ruben Zatini - 4 years, 3 months ago

What did you check for concavity? As $f{x} = \frac{1}{k + x}^2$ is convex as far as I know...

Vishal Yadav - 4 years, 3 months ago

$\frac{1}{(k +x)^2}$ I meant..

Vishal Yadav - 4 years, 3 months ago

Even better approach is to use the substitution $\displaystyle a + b = x, b + c = y, c + a = z$ . And then again using Jensen's inequality which can now easily be used giving $\displaystyle \le 3\frac{{3}^{2}}{4{S}^{2}}$ Remember here- $\displaystyle S = 2(a+b+c)$

And here again using the fact that $a + b + c \geq 3$

we get our answer as -

$\displaystyle \frac{3}{16}$

Kartik Sharma - 6 years, 3 months ago

Well, yeah it is almost the same thing but concavity here can more easily be proved. And yeah, @Parth Lohomi This problem too is only a Level 3 - 4 due to two reasons - 1. It is almost direct Jensen application and 2. It has the most common equality case( $(1,1,1)$ )

Kartik Sharma - 6 years, 3 months ago

Can you explain why the function is concave and not convex?

Zuxiang Kong - 10 months, 3 weeks ago

This problem is also there in Evan Chen's handout

Harry Jones - 4 years, 2 months ago

Which one?

Krishna Bhatraju - 2 years, 7 months ago

Could someone explain why is it concave please?

Chris Ye - 3 years, 7 months ago

Let $f(x) = \frac{1}{x}$ , which is convex.

$\frac{f(a) + f(b) + f(c)}{3} \geq f(\frac{a + b + c}{3})$

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq \frac{3*3}{a + b + c}$

Let $k = a + b + c = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq 0$ .

$k \geq \frac{9}{k}$

$k \geq 3$ is least when $a = b = c$ , i.e. when $k = 3$ and $a = b = c = 1$ .

Let $g(x) = x^{-2}$ , which is convex.

$(\frac{2a + b + c}{4})^{-2} \le \frac{2a^{-2} + b^{-2} + c^{-2}}{4}$

$(2a + b + c)^{-2} \le \frac{2a^{-2} + b^{-2} + c^{-2}}{64}$

$(a + 2b + c)^{-2} \le \frac{a^{-2} + 2b^{-2} + c^{-2}}{64}$

$(a + b + 2c)^{-2} \le \frac{a^{-2} + b^{-2} + 2c^{-2}}{64}$

Summing the above 3 inequalities, for any $k = a + b + c$ , $\dfrac{1}{\left(2a+b+c\right)^2}+\dfrac{1}{\left(2b+c+a\right)^2}+\dfrac{1}{\left(2c+b+a\right)^2} \le \frac{a^{-2} + b^{-2} + c^{-2}}{16}$ is maximized when $a = b = c$ .

The least possible $k = 3$ , where $a = b = c = 1$ , gives the greatest sum $= \frac{3}{16}$ .

Seq O - 2 years, 8 months ago

But when you take second derivative, it is positive.

Krishna Bhatraju - 2 years, 7 months ago

Divyansh Tripathi
Mar 16, 2015

LET A= B=C =1

Peter Macgregor
Mar 17, 2015

I correctly guessed the answer by noticing that the problem is completely symmetric in $(a,b,c)$ and so one might expect that the solution will have $a=b=c$ . Using this condition to write the constraint equation in terms of $a$ alone gives

$\dfrac{3}{a}=3a$ and so $a=b=c=1$

substituting these answers in the objective function gives

$\dfrac{3}{16}$ and so $m+n=\boxed{19}$

Of course this approach indicates little more than an educated guess. Here is a more convincing solution using the method of 'Lagrange multipliers for constrained optimisation'. Write

$H=\dfrac{1}{(2a+b+c)^2}+\dfrac{1}{(a+2b+c)^2}+\dfrac{1}{(a+b+2c)^2}+\lambda (\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-a-b-c)$

To optimise the objective function all four partial derivatives of $H$ must be zero. Differentiating with respect to $a$ yields

$\dfrac{-6}{(2a+b+c)^3}-\dfrac{\lambda}{a^2}-\lambda=0$

Differentiating H with respect to $b$ and $c$ yields two similar equations and differentiating H with respect to $\lambda$ recovers the constraint equation.

It is easy to check that these equations have solution

$a=b=c=1$

and (although it is not needed for the solution) $\lambda=\dfrac{-3}{64}$

Substituting $a=b=c=1$ into the objective function completes the problem.

Prashanth Subramanian
Mar 16, 2015

I am extremely sorry Karthik...........I just substituted and got the answer........! Sorry....! But what to do.......many like me...... without knowing how to solve......get the ans correct........you are great ....it's just that I can't understand..............! :p

I think I'm with Prashanth here. I used the first equation to postulate some more conditions and substituted these in to get the second equation in one variable.

The conditions I used were as follows: Symmetry about a,b, & c. One of the variables must be one, The remaining two must be recipricals.

As stated, I used these to derive the following:

c^2[(1/(c^2+2c+1)^2+1/(2c^2+c+1)^2+1/(c^2+c+2)^2

I saw the denominators would quickly outgrow the numerators as c grew large.

I suspect, though am not sure how to prove that there is just one one maxima.

So I used a spreadsheet to go through some examples and found c=1 was the larger than c=0 and that the function only seemed decreasing after that.

c=1 corresponds to 3/16 and the rest is covered above.

I'm still don't understand how to apply Jenson's Inequality to this. If my thinking about the shape of the functions graph is correct, the function is concave between any two points. So how do infinately many tests for concavity tell me anything about the maxima/minima of the function.

Carl White - 6 years ago

Maximising Can Also Be Difficult!

The answer is 19.

4 solutions

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