# Minimum of $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}$

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 $2$.

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.

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

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.

• Very nice answer!. – mfl Nov 2 '14 at 21:01

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!

• 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 $2$ – Umesh shankar Jan 31 '19 at 9:57
• @Umesh shankar The infimum is $2$ because our variables are positives. – Michael Rozenberg Jan 31 '19 at 10:36
• Yes i got it now – Umesh shankar Jan 31 '19 at 10:39