Find the maximum $P=\frac{bc}{\sqrt{(1-a)^3(a+1)}}+\frac{ca}{\sqrt{(1-b)^3(b+1)}}+\frac{ab}{\sqrt{(1-c)^3(c+1)}},$ when $a+b+c=1$

Question

Let $a,b,c>0: a+b+c=1.$ Find the maximum$$P=\frac{bc}{\sqrt{(1-a)^3(a+1)}}+\frac{ca}{\sqrt{(1-b)^3(b+1)}}+\frac{ab}{\sqrt{(1-c)^3(c+1)}}.$$ When $a=b=c=\dfrac{1}{3}$ P attained the value $\dfrac{3\sqrt{2}}{8}.$\

I tried to use AM-GM$$\frac{bc}{\sqrt{(1-a)^3(a+1)}}=\frac{bc}{(b+c)\sqrt{1-a^2}}\le \frac{\sqrt{bc}}{2\sqrt{1-a^2}} \le \frac{1-a}{4\sqrt{1-a^2}}=\frac{1}{4}.\sqrt{\frac{b+c}{2a+b+c} }.$$Thus, we need to prove $$\sqrt{\frac{a+b}{2c+a+b}}+\sqrt{\frac{c+b}{2a+c+b}}+\sqrt{\frac{a+c}{2b+a+a}}\le \frac{3\sqrt{2}}{2},$$which I don't know to prove it.

Hope to see some helps here. Thank you for interest.

Answer 1

By C-S and AM-GM we obtain: $$\sum_{cyc}\frac{bc}{\sqrt{(1-a)^3(a+1)}}=\sum_{cyc}\frac{bc}{\sqrt{(b+c)^3(2a+b+c)}}\leq$$ $$\leq\sqrt{\sum_{cyc}\frac{bc}{(b+c)(2a+b+c)}\sum_{cyc}\frac{bc}{(b+c)^2}}\leq\sqrt{\sum_{cyc}\frac{bc}{(b+c)(2a+b+c)}\cdot\frac{3}{4}}\leq$$ $$\leq\sqrt{\sum_{cyc}\frac{\sqrt{bc}}{2a+b+c}\cdot\frac{1}{2}\cdot\frac{3}{4}}\leq\sqrt{\frac{3}{8}\sum_{cyc}\frac{\sqrt{bc}}{2\sqrt{(a+b)(a+c)}}}\leq$$ $$\leq\sqrt{\frac{3}{32}\sum_{cyc}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)}=\sqrt{\frac{3}{32}\sum_{cyc}\left(\frac{b}{a+b}+\frac{a}{b+a}\right)}=\frac{3}{4\sqrt2}.$$

Answer 2

Proof.

According to your progress, it is enough to prove that$$\dfrac{bc}{\sqrt{(b+c)^3(b+c+2a)}}+\dfrac{ca}{\sqrt{(a+c)^3(a+c+2b)}}+\dfrac{ba}{\sqrt{(b+a)^3(b+a+2c)}} \le \dfrac{3\sqrt{2}}{8}.$$ We can apply AM-GM as \begin{align*} \dfrac{bc}{\sqrt{(b+c)^3(b+c+2a)}} &\le \dfrac{\sqrt{2}}{4}\left(\dfrac{bc}{(b+c)^2}+\dfrac{2bc}{(b+c)(b+c+2a)}\right)\\ &\le \dfrac{\sqrt{2}}{4}\left(\dfrac{1}{4}+\dfrac{bc}{(b+c)\sqrt{(a+b)(c+a)}}\right)\\ &\le \dfrac{\sqrt{2}}{4}\left(\dfrac{1}{4}+\dfrac{bc}{2(b+c)}\left(\dfrac{1}{a+b}+\dfrac{1}{c+a}\right)\right)\\ &=\dfrac{\sqrt{2}}{16}+\dfrac{\sqrt{2}}{8}\cdot\dfrac{ab(a+b+2c)}{(a+b)(b+c)(c+a)}. \end{align*} Similarly, we obtain$$\sum\limits_{cyc}\dfrac{bc}{\sqrt{(b+c)^3(b+c+2a)}} \le \dfrac{3\sqrt{2}}{16}+\dfrac{\sqrt{2}}{8}\left(1+\dfrac{4abc}{(a+b)(b+c)(c+a)}\right).$$ Id est, we will prove$$(a+b)(b+c)(c+a) \ge 8abc,$$which is equivalent to $$a(b-c)^2+b(c-a)^2+c(a-b)^2\ge 0.$$Thus, we got maximal value is equal to $\dfrac{3\sqrt{2}}{8}$ when $a=b=c=\dfrac{1}{3}.$