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

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.

**
This section requires Javascript.
**

You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.

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

Log in to reply

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

Log in to reply

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

Log in to reply

@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

Log in to reply

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

Priyanshu Mishra
- 5 years, 7 months ago

Log in to reply

@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

Log in to reply

@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

Log in to reply

@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

Log in to reply

@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

Log in to reply

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

Log in to reply

Which According to you are best inequalities?

Kushagra Sahni
- 5 years, 7 months ago

Log in to reply

@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

Log in to reply

@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

Log in to reply

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

Log in to reply

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

Otto Bretscher
- 5 years, 7 months ago

Log in to reply

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

Log in to reply

@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

Log in to reply

@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

Log in to reply

@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

Log in to reply

@Otto Bretscher – The problem asks to find max?

Dev Sharma
- 5 years, 7 months ago

Log in to reply

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

Otto Bretscher
- 5 years, 7 months ago

×

Problem Loading...

Note Loading...

Set Loading...

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 $a=b=c=\frac{1}{3}$ .