$\frac{3x+1}{x+1}+\frac{3y+1}{y+1}+\frac{3z+1}{z+1} \le \frac{9}{2}$

Question

I'm having trouble proving that for any $x,y,z>0$ such that $x+y+z=1$ the following inequality is true:

$\frac{3x+1}{x+1}+\frac{3y+1}{y+1}+\frac{3z+1}{z+1} \le \frac{9}{2}$

It seems to me that Jensen's inequality could do the trick, but I'm having trouble finding the right function and the right arguments. Any help is appreciated.

Answer 1

Hint: $f(x) = \dfrac{3x+1}{x+1}$. You then show $f”(x) < 0$.

Answer 2

Hint:

$$\left(\frac1{x+1}+\frac1{y+1}+\frac1{z+1}\right)(x+1+y+1+z+1) \ge 9$$

And

$$\frac{3x+1}{x+1} = 3 - \frac{2}{x+1}.$$

Answer 3

Another way.

We need to prove that $$\sum_{cyc}\left(\frac{3}{2}-\frac{3x+1}{x+1}\right)\geq0$$ or $$\sum_{cyc}\frac{1-3x}{x+1}\geq0$$ or $$\sum_{cyc}\left(\frac{1-3x}{x+1}+\frac{3}{4}(3x-1)\right)\geq0$$ or $$\sum_{cyc}(3x-1)\left(\frac{3}{4}-\frac{1}{x+1}\right)\geq0$$ or $$\sum_{cyc}\frac{(3x-1)^2}{x+1}\geq0$$ and we are done!