Given non-negative real numbers $a,b,c$ which sum is $2.$ Assume that $c=min\{a,b,c\}.$
Denote $$f(a,b,c)=\frac{\sqrt{2ab+7}+\sqrt{2bc+7}+\sqrt{2ca+7}}{\sqrt{abc+9}}.$$ Prove that $f(a,b,c)\le f(t,t,c),$ where $t=\dfrac{a+b}{2}.$
I've tried to use Cauchy inequality without success.
Indeed, we have
$$f\left(\frac{a+b}{2},\frac{a+b}{2},c\right)=\frac{\sqrt{\dfrac{(a+b)^2}{2}+7}+\sqrt{c(a+b)+7}+\sqrt{c(a+b)+7}}{\sqrt{\dfrac{(a+b)^2}{4}c+9}}.$$
Now, it's true by Cauchy that $\sqrt{2bc+7}+\sqrt{2ca+7}\le 2\sqrt{ac+bc+7} \iff \sqrt{2bc+7}.\sqrt{2ca+7}\le ac+bc+7. $
And $\sqrt{2ab+7}\le \sqrt{\dfrac{(a+b)^2}{2}+7}.$ Thus, it's enough to show $$\frac{1}{\sqrt{abc+9}}\le \frac{1}{\sqrt{\dfrac{(a+b)^2}{4}c+9}},$$which is equivalent to $(a+b)^2\le 4ab.$
It is reverse inequality. Hope you help me continue my thoughts. Thank you!
We need to prove that: $f(a,b,c)\geq0,$ where $$f(a,b,c)=\frac{2\sqrt7+3}{3}\sqrt{abc+9}-\sum_{cyc}\sqrt{2ab+7}.$$ Let $c=\min\{a,b,c\}.$
Thus, $0\leq c\leq\frac{2}{3}$ and by Jensen and AM-GM we obtain: $$f(a,b,c)-f\left(\frac{a+b}{2},\frac{a+b}{2},c\right)=-\frac{2\sqrt7+3}{3}\left(\sqrt{\left(\frac{a+b}{2}\right)^2c+9}-\sqrt{abc+9}\right)+$$ $$+\left(2\sqrt{(a+b)c+7}-\sqrt{2ac+7}-\sqrt{2bc+7}\right)+\left(\sqrt{\frac{(a+b)^2}{2}+7}-\sqrt{2ab+7}\right)=$$ $$=-\tfrac{\frac{2\sqrt7+3}{3}\cdot \frac{(a-b)^2c}{4}}{\sqrt{\left(\frac{a+b}{2}\right)^2c+9}+\sqrt{abc+9}}+\tfrac{\left(\sqrt{2ac+7}-\sqrt{2bc+7}\right)^2}{2\sqrt{(a+b)c+7}+\sqrt{2ac+7}+\sqrt{2bc+7}}+\tfrac{\frac{(a-b)^2}{2}}{\left(\sqrt{\frac{(a+b)^2}{2}+7}+\sqrt{2ab+7}\right)}=$$ $$=(a-b)^2\left(-\tfrac{(2\sqrt7+3)c}{12\left(\sqrt{\left(\frac{a+b}{2}\right)^2c+9}+\sqrt{abc+9}\right)}+\tfrac{4c^2}{\left(2\sqrt{(a+b)c+7}+\sqrt{2ac+7}+\sqrt{2bc+7}\right)\left(\sqrt{2ac+7}+\sqrt{2bc+7}\right)^2}+\tfrac{1}{2\left(\sqrt{\frac{(a+b)^2}{2}+7}+\sqrt{2ab+7}\right)}\right)\geq$$ $$\geq(a-b)^2\left(-\tfrac{(2\sqrt7+3)c}{12\left(\sqrt{\left(\frac{2-c}{2}\right)^2c+9}+\sqrt{c^3+9}\right)}+\tfrac{4c^2}{4\sqrt{(2-c)c+7}\left(2\sqrt{(2-c)c+7}\right)^2}+\tfrac{1}{4\sqrt{\frac{(2-c)^2}{2}+7}}\right).$$ Can you end it now?