Let $a;b;c>0$ such that $ab+bc+ca=1$. Find the maximize value of $$K=\sqrt{\frac{a}{a+c}}+\sqrt{\frac{b}{b+c}}-\frac{9\sqrt{c^2+1}}{8c}$$
I can see: $$a=b=\frac{1}{\sqrt{7}};c=\frac{3}{\sqrt{7}}\rightarrow K=\frac{-1}{2}$$ so i will prove $K=\frac{-1}{2}$ is maximize value of $K$.
Indeed, i will prove $\sqrt{\frac{a}{a+c}}+\sqrt{\frac{b}{b+c}}-\frac{9\sqrt{c^2+1}}{8c}\le -\frac{1}{2}$
Or $\sqrt{\frac{a}{a+c}}-\frac{1}{2}+\sqrt{\frac{b}{b+c}}-\frac{1}{2}-\left(\frac{9\sqrt{c^2+ab+bc+ca}}{8c}-\frac{3}{2}\right)\le 0$
Or $$\frac{\frac{3a-c}{4a+c}}{\sqrt{\frac{a}{a+c}}+\frac{1}{2}}+\frac{\frac{3b-c}{b+c}}{\sqrt{\frac{b}{b+c}}+\frac{1}{2}}-\frac{3}{2}\cdot \frac{\frac{-7c^2+9ab+9bc+9ca}{3\sqrt{c^2+ab+bc+ca}+4c}}{4c}\le 0$$
Or $$\frac{\frac{3a-c}{4a+c}}{\sqrt{\frac{a}{a+c}}+\frac{1}{2}}+\frac{\frac{3b-c}{b+c}}{\sqrt{\frac{b}{b+c}}+\frac{1}{2}}-\frac{3}{2}\cdot \frac{\frac{9ab-c^2+3c\left(3a-c\right)+3c\left(3b-c\right)}{3\sqrt{c^2+ab+bc+ca}+4c}}{4c}\le 0$$
Then we have: $9ab\le \frac{\left(a+b\right)^2}{4}\Leftrightarrow 9ab-c^2\le \frac{9\left(a+b\right)^2}{4}-c^2=\left(\frac{3a-c+3b-c}{2}\right)\left(\frac{3a+3b}{2}+c\right)$
Or $$\frac{\frac{3a-c}{4\left(a+c\right)}}{\sqrt{\frac{a}{a+c}}+\frac{1}{2}}+\frac{\frac{3b-c}{4\left(b+c\right)}}{\sqrt{\frac{b}{b+c}}+\frac{1}{2}}-\frac{3}{2}\frac{\frac{\left(3a-c\right)\cdot \frac{1}{2}\cdot \left(\frac{3a+3b}{2}+c\right)+\left(3b-c\right)\cdot \frac{1}{2}\cdot \left(\frac{3a+3b}{2}+c\right)+3c\left(3b-c\right)+3c\left(3a-c\right)}{3\sqrt{c^2+ab+bc+ca}+4c}}{4c}\le 0$$
It's easy to recognize the equality occurs at $3a=3b=c$ so i am trying to take it becomes $(3a-c)^2\cdot f(a;b;c)+(3b-c)^2\cdot f(a;b;c)\ge 0$ but i do not know how to do it. Help me !!
P/s:This problem was extended by Prove that $2\sqrt{\frac{a}{a+c}}-\frac{9\sqrt{c^2+1}}{8c}<1$ with $a;c>0$
By your work we need to prove that $$\sqrt{\frac{a}{a+c}}+\sqrt{\frac{b}{b+c}}-\frac{9\sqrt{(a+c)(b+c)}}{8c}\leq-\frac{1}{2}.$$ Now, let $\frac{a}{a+c}=\frac{x^2}{4}$ and $\frac{b}{b+c}=\frac{y^2}{4},$ where $x$ and $y$ be positives.
Thus, $0<x<2$ and $0<y<2$ and since $$\frac{a+c}{c}=\frac{1}{\frac{c}{a+c}}=\frac{1}{1-\frac{a}{a+c}}=\frac{1}{1-\frac{x^2}{4}}=\frac{4}{4-x^2},$$ we need to prove that $$\frac{x}{2}+\frac{y}{2}-\frac{9}{2\sqrt{(4-x^2)(4-y^2)}}\leq-\frac{1}{2}$$ or $$(x+y+1)\sqrt{(4-x^2)(4-y^2)}\leq9$$ or $$\left(\frac{x+y+1}{3}\right)^2\cdot\frac{2+x}{3}\cdot\frac{2+y}{3}\cdot(2-x)\cdot(2-y)\leq1,$$ which is true by AM-GM: $$\left(\frac{x+y+1}{3}\right)^2\cdot\frac{2+x}{3}\cdot\frac{2+y}{3}\cdot(2-x)\cdot(2-y)\leq$$ $$\leq\left(\frac{2\cdot\frac{x+y+1}{3}+\frac{2+x}{3}+\frac{2+y}{3}+2-x+2-y}{6}\right)^6=1.$$