For any $a,b,c\ge 0$ prove that $$3a+4(b+c)+\sqrt[3]{abc}\ge 2\left( \sqrt{b\left(4b+5a\right) }+ \sqrt{c\left(4c+5a\right) }\right).$$
Equality holds at $a=b=c=1$ or $a=0$
I tried to use AM as $$6\sqrt{b\left(4b+5a\right) }=2\sqrt{9b\left(4b+5a\right) }\le 13b+5a$$
$$6\sqrt{c\left(4c+5a\right) }=2\sqrt{9c\left(4c+5a\right) }\le 13c+5a$$
and we need to prove $$9a+12(b+c)+3\sqrt[3]{abc}\ge 13b+5a+13c+5a$$
or $$3\sqrt[3]{abc}\ge a+b+c$$which is reverse since by AM-GM $3\sqrt[3]{abc}\le a+b+c.$
Is there a better approach? Any ideas and comments is welcome.
Best Answer
This is not an answer, might be too long for a comment. It is definitely not elegant but probably doable if one tries hard. : )
Let us assume $a > 0$ otherwise the equality holds, and let $u=\frac{b}{a}$, $v=\frac{c}{a}$, then the problem is equivalent to
$$f(u,v) := 3 + 4(u+ v) + u^{1/3} v^{1/3} - 2 u^{1/2} (4u+5)^{1/2} - 2 v^{1/2} (4v+5)^{1/2}\ge 0$$
and one can exclude the possibility of a local minimizer on the boundary, then solve
$$f_u = \frac{1}{3} u^{-2/3} v^{1/3} - \frac{(2\sqrt{u} - \sqrt{4u+5})^2}{\sqrt{u(4u+5)}} = 0$$
$$f_v = \frac{1}{3} v^{-2/3} u^{1/3} - \frac{(2\sqrt{v} - \sqrt{4v+5})^2}{\sqrt{v(4v+5)}} = 0$$
which gives $v = F(u), u = F(v)$, where $$F(x) =27 \left( \frac{(2\sqrt{x} - \sqrt{4x+5})^2}{\sqrt{x(4x+5)}} \right)^3 x^{2} = 27\left( \frac{25}{\sqrt{x(4x+5)}(2\sqrt{x} + \sqrt{4x+5})^2} \right)^3 x^{2}=27 \left( \frac{25}{\sqrt{(4x+5)}(2\sqrt{x} + \sqrt{4x+5})^2} \right)^3 x^{1/2},$$ then all we need is the fixed point of $F\circ F$, it seems there is no quick way for that.