Suppose that $$f(x) = \frac{e^x-1} {x(x+a)},$$ where $x>0$ and $a>0$, how to prove $f$ is convex in $x$? If it is not convex, do you have a counterexample?
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1For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – Although, as a 3 year member of this site, you should know that :) – Martin R Nov 11 '21 at 03:18
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1@coffeemath fixed. – Adam Nov 11 '21 at 03:25
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1it's not that bad. you just have to simplify the first derivative and you only have to differentiate the numerator of $f'(x)$ to get the sign of $f''(x)$. You can make use of the fact that the product of two monotonically increasing and continuous functions is also monotonically increasing for $x \geq 0$ – okzoomer Nov 11 '21 at 03:26
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"you only have to differentiate the numerator of f′(x) to get the sign of f′′(x)." Care to elaborate, @okzoomer? I am not very good at math. – Adam Nov 11 '21 at 03:34
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$\displaystyle f‘(x) = \frac{e^x(x^2+ax) - (e^x-1)(2x+a)}{x^2(x+a)^2} = \frac{e^x(x^2+(a-2)x - a) + 2x + a}{x^2(x+a)^2}$ the denominator is always non-negative – okzoomer Nov 11 '21 at 03:40
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@okzoomer I wish I understood what you wrote, but I don't. How does it prove convexity? – Adam Nov 11 '21 at 03:49
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@Adam I take back what I said about only differentiating the numerator. It has more stipulations than what I mentioned. Instead you should look to simplify by writing out $x(x+a)$ as $x^2 + ax$ and $x^2(x+a)^2$ as $x^4 + 2ax^3 + a^2x^2$ and proceed. – okzoomer Nov 11 '21 at 04:10
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Let us continue this discussion in chat. – Adam Nov 11 '21 at 04:35
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Use brute force and group terms to have $$f''(x)=\frac{A(x)\,a^2+B(x)\,a+C(x)}{x^3 (x+a)^3}$$ with $$A(x)=e^x (x^2-2 x+2)-2$$$$B(x)=2x(e^x(x^2-3 x+3)-3)$$ $$C(x)=x^2( e^x(x^2-4 x+6)-6)$$ Show that these three terms are always positive for $x>0$.
Compute $\Delta=B^2(x)-4A(x)C(x)$ and show that is it always negative. If this is true, then $f''(x)>0 \quad \forall (a,x)$.
Claude Leibovici
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