The Vitals of Mathematics

Algebra Level 5

$a+b+c=\lambda$ for positive real variables $a,b,c$ .
Let $S_1=\sum \limits_{cyc} \frac{a^2}{\lambda-a}$ and $S_2=\sum \limits_{cyc} \frac{1}{\lambda-a}$ .
We define a function $f(t)$ such that $f(S_2)=S_1$ . If $f'(t) = 9$ for all values of $t$ in the domain, determine the value of $\lambda$ .

Note: This is my original problem belonging to the set Questions I've Made © . Give a try to its sister problem

The answer is 3.

1 solution

Sanjeet Raria
Feb 17, 2015

One idea to approach this question is to express $S_1$ in terms of $S_2$ so as to use the notion of derivative later. Now we have $S_2=\sum\limits_{cyc} \frac{1}{\lambda-a}=\sum\limits_{cyc} \frac{1}{b+c}=t(say)$ and $S_1=\sum\limits_{cyc} \frac{a^2}{\lambda-a} =\sum\limits_{cyc} \frac{a^2}{b+c}=f(t)(say)$

Now the next part is to find $\large f(t)$ . Here's the direct result I'm gonna write. Better practice it yourself: $\sum\limits_{cyc} \frac{a^2}{b+c}=(a+b+c)^2 \sum\limits_{cyc} \frac{1}{b+c}-4(a+b+c)$ So we have $\large f(t)={\lambda}^2•t-4\lambda$

Now according to the question, Mathematically we can write, $\frac{df(t)}{dt}=9$ $\rightarrow f'(t)= {\lambda}^2=9$ $\Rightarrow \lambda=\boxed 3$ Since $a,b,c$ are positive reals. Other approaches are highly welcomed

Good solution I used a slightly different method I don't know if it is correct.

I used Cauchy Schwartz inequality to find the minimum value of $S_{1} and S_{2}$

Then I equated them to find that $f(t)= \lambda ^{2} \times t$

Then it was quite simple.

I guess I got lucky

Aayush Patni - 6 years, 3 months ago

Sir it is a very nice question!Please post more questions like this!

@Sanjeet Raria

Harsh Shrivastava - 6 years, 1 month ago

Thank you Harsh.

Sanjeet Raria - 6 years, 1 month ago

The Vitals of Mathematics

Note: This is my original problem belonging to the set Questions I've Made © . Give a try to its sister problem

The answer is 3.

1 solution

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