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Clearly when $z=2$ we have $z = 2(z+1-\log z) - (4-2\log 2) $.

The slope when $z>2$ is greater than $1$ so clearly the inequality is true on $[2,\infty)$

How do you show it still holds true on $(0,2)$? I don't see any trivial reason why it's true (plotting it shows it's true so I know it's true, but I'd like an analytical reason for why). Rearranging the RHS I get:

$$2z-2 + 2\log(\frac{2}{z})$$

But I don't see why $z$ is less than that on $(0,2)$

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Hint: The RHS is convex and the LHS is the tangent line at $z=2$.

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Consider the function $$f(z)= 2(z+1-\log z) - (4-2\log 2)-z$$ for which differentiation gives $$f'(z)=1-\frac 2z$$ $$f''(z)=\frac 2{z^2}$$ So, the first derivative cancels for $z=2$; at this point $f(2)=0$ and the second derivative test shows that this is then a minimum. So, the inequality holds.

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