10
$\begingroup$

What is the minimum of $$f(a,b,c):=\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}$$ where $a,b,c$ are positive real numbers?

When $a=b=c$, we have $f(a,b,c)=\dfrac{3}{\sqrt{2}}\approx 2.12$

When $a=1,b=c\rightarrow\infty$, we have $f(a,b,c)\rightarrow 2$. So the minimum is at most $2$.

$\endgroup$
6
$\begingroup$

Hint: this is also $$ \min_{a,b,c\ge 0, a+b+c=1} \sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}} =\min_{a,b,c\ge 0, a+b+c=1} \sqrt{\dfrac{a}{1-a}}+\sqrt{\dfrac{b}{1-b}}+\sqrt{\dfrac{c}{1-c}} $$And then you can for instance use the Lagrange multiplicators.

$\endgroup$
  • $\begingroup$ Where did the constraint $a + b + c = 1$ come from? $\endgroup$ – Simon S Nov 2 '14 at 20:36
  • $\begingroup$ @SimonS this is to simplify the denominator. $\endgroup$ – mookid Nov 2 '14 at 20:38
  • 1
    $\begingroup$ You can renormalize all 3 numbers by the same factor without changing the function. so $f(a,b,c)=f(a/(a+b+c),b/(a+b+c),c/(a+b+c))$. $\endgroup$ – Clement C. Nov 2 '14 at 20:38
  • $\begingroup$ Very nice hint! We can also avoid the use of Lagrange multiplicators, as I just posted. $\endgroup$ – simmons Nov 2 '14 at 20:48
  • $\begingroup$ @SimonS both solutions use the renormalisation. This is independent to the validity of Lagrange multipliers. $\endgroup$ – mookid Nov 2 '14 at 20:51
9
$\begingroup$

Following mookid's hint, we can also avoid the use of Lagrange multiplicators. Normalize so that $a+b+c=1$, and then use the inequality $\sqrt{\dfrac{a}{1-a}}\geq 2a$. This is equivalent to $a(2a-1)^2\geq 0$.

Hence $f(a,b,c)\geq 2(a+b+c)=2$. Equality cannot hold, since $a=b=c=\dfrac{1}{2}$ doesn't satisfy $a+b+c=1$. But $f(a,b,c)$ can arbitrary get close to $2$, as the example in the original question shows.

$\endgroup$
  • $\begingroup$ Very nice answer!. $\endgroup$ – mfl Nov 2 '14 at 21:01
3
$\begingroup$

If $a=b=1$ and $c\rightarrow0^+$ we get $f(a,b,c)\rightarrow2$.

We'll prove that $$\sum_{cyc}\sqrt{\frac{a}{b+c}}\geq2.$$ Indeed, by AM-GM $$\sum_{cyc}\sqrt{\frac{a}{b+c}}=\sum_{cyc}\frac{2a}{2\sqrt{a(b+c)}}\geq\sum_{cyc}\frac{2a}{a+b+c}=2.$$ Done!

Also we can use Holder: $$\left( \sum_{cyc}\sqrt{\frac{a}{b+c}}\right)^2\sum_{cyc}a^2(b+c)\geq(a+b+c)^3.$$ Thus, it remains to prove that $$(a+b+c)^3\geq4\sum_{cyc}(a^2b+a^2c)$$ or $$\sum_{cyc}(a^3-a^2b-a^2c+2abc)\geq0,$$ which follows from Schur.

Done!

$\endgroup$
  • $\begingroup$ Very nice, so you mean to say we cannot find exact value of $c$ when $a=b=1$ such that $f(a,b,c)=2$ but rather we can get a limit as $2$ $\endgroup$ – Umesh shankar Jan 31 at 9:57
  • $\begingroup$ @Umesh shankar The infimum is $2$ because our variables are positives. $\endgroup$ – Michael Rozenberg Jan 31 at 10:36
  • $\begingroup$ Yes i got it now $\endgroup$ – Umesh shankar Jan 31 at 10:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.