Let $a\ge b\ge c\ge 0$ such that $a+b+c=1$ Find the minimum value of $P=\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\dfrac{24}{5\sqrt{5a+5b}}$

I found that the minimum value of $P$ is $\dfrac{78}{5\sqrt{15}}$ when $a=b=\dfrac{3}{8};c=\dfrac{1}{4}$

And this is my try

Applying AM-GM inequality, we get: $\dfrac{b+c}{a}+\dfrac{5}{3}\ge2\sqrt{\dfrac{5}{3}}.\sqrt{\dfrac{b+c}{a}}$

This implies $\sqrt{\dfrac{a}{b+c}}\ge2\sqrt{\dfrac{5}{3}}.\dfrac{3a}{3+2a}$

Similarly, $\sqrt{\dfrac{b}{a+c}}\ge2\sqrt{\dfrac{5}{3}}.\dfrac{3b}{3+2b}$

We need to prove that: $2\sqrt{\dfrac{5}{3}}\left(\dfrac{3a}{3+2a}+\dfrac{3b}{3+2b}\right)+\dfrac{24}{5\sqrt{5a+5b}}\ge\dfrac{78}{5\sqrt{15}}$

But I have no idea how to continue. Who can help me or have any other idea?


2 Answers 2



$m=\sqrt{\dfrac{a}{1-a}},n=\sqrt{\dfrac{b}{1-b}} \implies a=\dfrac{m^2}{m^2+1},b=\dfrac{n^2}{n^2+1},\\a\ge b\ge c \implies a\ge \dfrac{1}{3} \implies m^2\ge \dfrac{1}{2} ,a+b=1-c \ge 1-b \implies n\ge \sqrt{\dfrac{1}{2m^2+1}}\implies mn\ge \sqrt{\dfrac{m^2}{2m^2+1}}=\sqrt{\dfrac{1}{2+\dfrac{1}{m^2}}}\ge \dfrac{1}{2} \implies 4mn \ge 2 $

$\dfrac{1}{a+b}=\dfrac{m^2n^2+m^2+n^2+1}{2m^2n^2+m^2+n^2}\ge \dfrac{(\dfrac{m+n}{2})^2+1}{2(\dfrac{m+n}{2})^2} \iff (n-m)^2(n^2+m^2+4mn-2) \ge 0$

$(\dfrac{m+n}{2}) =\dfrac{1}{t} \implies P \ge \dfrac{2}{t}+\dfrac{24}{5\sqrt{10}}\sqrt{1+t^2}=2f(t) \\ f'(t)=-\dfrac{1}{t^2}+\dfrac{12}{5\sqrt{10}}\sqrt{\dfrac{t^2}{1+t^2}}=0 \implies t^2=\dfrac{5}{3}$

it is easy to verify this is the min point.

$f_{min}=\dfrac{39}{5\sqrt{15}}, P_{min}=\dfrac{78}{5\sqrt{15}}$ when $m=n$

so $m=n=\dfrac{1}{t} \implies a=b=\dfrac{1}{1+t^2}=\dfrac{3}{8}>\dfrac{1}{3}$


You can use calculus to find this.

$f(a,b) = \sqrt{\frac{a}{1-a}} + \sqrt{\frac{b}{1-b}} + \frac{24}{5\sqrt{5(a + b)}}$

Now, let us first find the critical point, $(x,y)$, of this function, where $f_a(x,y) = 0$ and $f_b(x,y) = 0$.

Taking partial derivatives and setting them to 0, you will see that the critical point is at $a = b$.

So we now have a 1D function, $g(a) = f(a,a) = 2\sqrt{\frac{a}{1 - a}}+ \frac{24}{5\sqrt{10 a}}$.

Then differentiate $g$ with respect to $a$, set it equal to 0, and you get that $a = 3/8$, so then $b = 3/8$ and $c = 1/4$.

You can make this more concrete by constructing the Hessian matrix of $f(a,b)$ and verifying that this is indeed the minimum (check that the determinant is >0 and that $f_{aa} > 0$ at this critical point).


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