199 upvotes problem

Algebra Level 4

$\large \dfrac{1}{a^{2}+9} + \dfrac{1}{b^{2}+9} + \dfrac{1}{c^{2}+9}$

Given that $a,b$ and $c$ are positive reals with a sum of 1. If the maximum value of the above expression can be expressed as $\dfrac{x}{y}$ where $x$ and $y$ are coprime integers, find the value of $x+y$ .

The answer is 109.

1 solution

Otto Bretscher
Oct 20, 2015

The function $f(x)=\frac{1}{x^2+9}$ is concave (also called "concave down") for $|x|\leq \sqrt{3}$ since $f''(x)=\frac{6(x^2-3)}{(x^2+9)^3}$ . Thus, by Jensen's Inequality , $f(a)+f(b)+f(c)\leq 3 f\left(\frac{a+b+c}{3}\right)=3f\left(\frac{1}{3}\right)=\frac{27}{82}$ , and $x+y=\boxed{109}$ . Equality holds when $a=b=c=\frac{1}{3}$ .

Sir, I've a doubt. I've solved many inequalities on brilliant using symmetry. Y does it work all the time? In this question I took symmetry for a,b,c. I got the idea for symmetry by inspiring from calculus. E.g. for a given perimeter of a triangle, the minimum area is that of an equilateral triangle. That's y I use a=b=c in inequalities. Is there a theoretical proof for this?

Aditya Kumar - 5 years, 7 months ago

The extrema are often attained when the variables are equal, but not always. You can't count on it, and you still have to examine each case separately. I will post a simple problem to illustrate this point.

Otto Bretscher - 5 years, 7 months ago

Sir, is this not correct?

By AM-GM

${ a }^{ 2 }+9\ge 6a$

So, $\large\ \frac { 1 }{ { a }^{ 2 }+9 } \le \frac { 1 }{ 6a }$ .

Similarly, $\large\ \frac { 1 }{ b^{ 2 }+9 } \le \frac { 1 }{ 6b }$ and

$\large\ \frac { 1 }{ c^{ 2 }+9 } \le \frac { 1 }{ 6c }$ .

Adding them yields

$\large\ \frac { 1 }{ a^{ 2 }+9 } +\frac { 1 }{ b^{ 2 }+9 } +\frac { 1 }{ c^{ 2 }+9 } \le \frac { 1 }{ 6a } +\frac { 1 }{ 6b } +\frac { 1 }{ 6c }$

The R.H.S is equivalent to

$\large\ \frac { 1 }{ 6 } \left( \frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } \right)$

And by AM-HM

$\large\ \frac { a+b+c }{ 3 } \ge \frac { 3 }{ \frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } }$ .

i.e, $\large\ \frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } \le \frac { 9 }{ a+b+c } =9$

Hence, the expression given in the question is less than or equal to $\large\ \frac { 9 }{ 6 }$ or $\large\ \frac { 3 }{ 2 }$ .

Hence $x + y = 3 + 2 = \boxed{5}$ .

I know that $5$ is not the answer , but sir please tell me where am i wrong?

Priyanshu Mishra - 5 years, 7 months ago

@Priyanshu Mishra – It should be $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \ge 9$ not $\le$ .

A Former Brilliant Member - 5 years, 7 months ago

@Priyanshu Mishra – Svatejas points out one mistake in your work. Also, showing that $\frac{3}{2}$ is an upper bound does not show that it is the maximum... the value is not attained.

Otto Bretscher - 5 years, 7 months ago

@Otto Bretscher – OK. So, there do not exist AM-GM method to solve this problem?

Priyanshu Mishra - 5 years, 7 months ago

@Priyanshu Mishra – I'm no expert and no fan of "AM-GM"... I don't think I have ever used the technique in my life. As you know, AM-GM is a special case of Jensen's inequality anyway.

Otto Bretscher - 5 years, 7 months ago

@Priyanshu Mishra – It s absolutely right as I see here. The am hm method is fine and 5 is a suitable answer

Aditya Narayan Sharma - 5 years, 6 months ago

Sir, in those kind of problems, I take x=y=0 z=1. That has also worked. I feel all these relations can be proved using calculus.

Aditya Kumar - 5 years, 7 months ago

@Aditya Kumar – With this simple example, I just wanted to show you that the solution isn't always $a=b=c$ when the equations are symmetric.

Sure, you can solve these problems with calculus... Lagrange multipliers are a powerful tool.

Otto Bretscher - 5 years, 7 months ago

@Otto Bretscher – how and when to use lagrange multiplier please explain in detail and what are its conditions

Akshay Sharma - 5 years, 7 months ago

@Akshay Sharma – You want the partial derivatives of the objective function $f$ to be proportional to the partial derivatives of the constraint $g=0$ , meaning that $f_a=\lambda g_a, f_b=\lambda g_b, f_c=\lambda g_c$ for some $\lambda$ . In this case we have $g_a=g_b=g_c=1$ , so that the partial derivatives of $f$ are required to be equal.

Otto Bretscher - 5 years, 7 months ago

This is because of equality condition in AM>=GM

Aakash Khandelwal - 5 years, 7 months ago

Can u solve it by am>=hm and am of mth power >=mth power of am? I am having a bit problem.

Shyambhu Mukherjee - 5 years, 7 months ago

Maybe you can... I'm not very fond of these AM-GM-HM type of inequalities. They seem too limited in scope, and their application almost always seems forced and contrived.

Otto Bretscher - 5 years, 7 months ago

Which According to you are best inequalities?

Kushagra Sahni - 5 years, 7 months ago

@Kushagra Sahni – In 40 years of experience in math (as a student, teacher, author, and researcher), I have never used an AM-GM-HM type of inequality, and I have never seen a mathematician use those (in seminar talks and such). The most widely used inequality is Cauchy-Schwarz, with Hoelder and Jensen being used occasionally.

Mathematicians usually use Lagrange multipliers or other calculus techniques to solve constraint optimization problems.

Otto Bretscher - 5 years, 7 months ago

@Otto Bretscher – Yeah I also love Cauchy-Schwarz.

Kushagra Sahni - 5 years, 7 months ago

In reponse to Otto Brestscher-Sir I want to know more about inequalities in calculus...can you suggest me some references or papers..I am a beginner in calculus and I am loving this subject..a direction is needed..hope u can help

Righved K - 5 years, 7 months ago

I think the most important technique in constrained optimization is Lagrange multipliers... there is a good lecture online from MIT . Mathematicians rarely use inequalities in this context... if so, mostly Cauchy-Schwarz and perhaps Jensen's.

Otto Bretscher - 5 years, 7 months ago

@Otto Bretscher u have only shown that the maximum value the given expression can obtain when $0 \le a,b,c \le \sqrt{3}$ is $\frac{27}{82}$ . Do you have a proof that an yet bigger value cannot be achieved for $\sqrt{3} \le a,b,c \le 1$ ?

A Former Brilliant Member - 3 years, 7 months ago

Its direct application of Titu's Lemma.

Dev Sharma - 5 years, 7 months ago

I don't think so....show us!

Otto Bretscher - 5 years, 7 months ago

it would be (1 + 1 + 1)^3 / (a^2 + b^2 + c^2 + 27)

now the value of a^2 + b^2 + c^2 can be easily get by QM AM.

Putting values we got it!!

Dev Sharma - 5 years, 7 months ago

@Dev Sharma – Forgive me for being a bit slow, but I don't see where this is going, at all. You promised us a direct application of Titu's lemma ;) Can you spell it out for us?

Otto Bretscher - 5 years, 7 months ago

@Otto Bretscher – yes, my first step is by Titu's Lemma, which is a generalization of Cauchy Scharwz

Dev Sharma - 5 years, 7 months ago

@Dev Sharma – But how does that help with our problem? I think it doesn't... (by the way, Titu's lemma is not a generalization of Cauchy Schwarz, but a very very special case)

Otto Bretscher - 5 years, 7 months ago

@Otto Bretscher – The problem asks to find max?

Dev Sharma - 5 years, 7 months ago

@Dev Sharma – so show us how you find that max.

Otto Bretscher - 5 years, 7 months ago

199 upvotes problem

The answer is 109.

1 solution

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