5
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

$P=\sqrt{\dfrac{a}{1-a}}+\sqrt{\dfrac{b}{1-b}}+\dfrac{24}{5\sqrt{5a+5b}}$

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

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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).

| cite | improve this answer | |
$\endgroup$

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.