For $a,b,c\ge0: ab+bc+ca+abc=4$ then: $$\frac{2-a}{a+\sqrt{bc+abc}}+\frac{2-b}{b+\sqrt{ca+abc}}+\frac{2-c}{c+\sqrt{ab+abc}}\ge1$$
I used the condition and get: $a+\sqrt{bc+abc}=a+\sqrt{4-a(b+c)}\le a+2$
So we need to prove that: $$\frac{2-a}{a+2}+\frac{2-b}{b+2}+\frac{2-c}{c+2}\ge1$$
I tried to full expand but the rest seems complicated for me.
Can anyone help me full my idea? Every thinking is welcomed, thanks!
Best Answer
Let $x = a + 2, \; y = b + 2, \; z = c + 2$
Hence, we need to prove that $$\frac{4-x}{x} + \frac{4-y}{y} + \frac{4-z}{z} \geq 1$$ $$\implies \frac{4}{x} - \frac{x}{x} + \frac{4}{y} - \frac{y}{y} + \frac{4}{z} - \frac{z}{z} \geq 1$$ $$\implies \frac{4}{x} + \frac{4}{y} + \frac{4}{z} - 3 \geq 1$$ $$\implies \frac{4}{x} + \frac{4}{y} + \frac{4}{z} \geq 4$$ $$\implies \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq 1$$ $$\implies \frac{1}{a+2} + \frac{1}{b+2} + \frac{1}{c+2} \geq 1$$ $$\implies \frac{4(a + b + c) + ab + bc + ca + 12}{4(a + b + c) + 2(ab + bc + ca) + abc + 8} \geq 1$$ $$\implies 4(a + b + c) + ab + bc + ca + 12 \geq 4(a + b + c) + 2(ab + bc + ca) + abc + 8$$ $$\implies ab + bc + ca +abc -4 \leq 0$$
We know that $ab + bc + ca + abc = 4 \implies ab + bc + ca +abc - 4 = 4 - 4 = 0$.