# Solve inequality

Given positive numbers $$a,b,c$$ satisfying $$a^2+b^2+c^2=1$$, prove the following inequality $$\frac{a}{\sqrt{1-bc}} + \frac{b}{\sqrt{1-ac}}+\frac{c}{\sqrt{1-ab}}\le\frac{3}{\sqrt{2}}$$ Thanks

I have tried using CS, try to make use of $$a+b+c\leq\sqrt3$$, $$abc\leq\frac{1}{3\sqrt3}$$, but got nowhere –

• What have you tried? Jul 23, 2019 at 2:00
• I have tried using CS, try to make use of $a+b+c\le\sqrt{3}$, $abc\le\frac{1}{3\sqrt{3}}$, but got nowhere Jul 23, 2019 at 2:13

$$\dfrac{a}{\sqrt{2-2bc}} \le \dfrac{a}{\sqrt{1+a^2}}$$
if you can prove $$f(x)=\sqrt{\dfrac{x}{1+x}}$$ is concave function, then the problem is solved.
• I am not sure the twice-differentiate is the solid way. The more safe way is use $f(\frac{{x+y}{2}}) \ge \frac{{f(x)+f(y)}{2}}$ Jul 30, 2019 at 1:08
By C-S $$\sum_{cyc}\frac{a}{\sqrt{1-bc}}\leq\sqrt{\sum_{cyc}a\sum_{cyc}\frac{a}{1-bc}}.$$ Thus, it's enough to prove that $$\sum_{cyc}\frac{a}{1-bc}\leq\frac{9}{2(a+b+c)},$$ which is true by SOS: $$\frac{9}{2(a+b+c)}-\sum_{cyc}\frac{a}{1-bc}=\sum_{cyc}\left(\frac{3}{2(a+b+c)}-\frac{a}{1-bc}\right)=$$ $$=\frac{1}{2(a+b+c)}\sum_{cyc}\frac{3(a^2+b^2+c^2-bc)-2a(a+b+c)}{1-bc}=$$ $$=\frac{1}{4(a+b+c)}\sum_{cyc}\frac{2a^2+6b^2+6c^2-6bc-4ab-4ac}{1-bc}=$$ $$=\frac{1}{4(a+b+c)}\sum_{cyc}\frac{(c-a)(6c-a-3b)-(a-b)(6b-a-3c)}{1-bc}=$$ $$=\frac{1}{4(a+b+c)}\sum_{cyc}(a-b)\left(\frac{6a-b-3c}{1-ac}-\frac{6b-a-3c}{1-bc}\right)=$$ $$=\frac{1}{4(a+b+c)}\sum_{cyc}\frac{(a-b)^2(7-c(a+b)-3c^2)}{(1-ac)(1-bc)}=$$ $$=\frac{1}{4(a+b+c)}\sum_{cyc}\frac{(a-b)^2(4c^2-c(a+b)+7(a^2+b^2))}{(1-ac)(1-bc)}=$$ $$=\frac{1}{4(a+b+c)}\sum_{cyc}\frac{(a-b)^2\left(\left(2c-\frac{a+b}{4}\right)^2+7(a^2+b^2)-\frac{(a+b)^2}{16}\right)}{(1-ac)(1-bc)}\geq0.$$