Show that $h(e)

Question

Problem :

Define for $a=0.5$ and $x=e$:

$$f(x)=\left(1+\frac{1}{x+a}\left(\frac{4\left(x+a\right)^{2}}{\left(x+a+2\right)^{2}}-1\right)\right)\left(x+a-\frac{2\left(x+a\right)}{\left(x+a+2\right)}\right)-\left(ax\right)^{\frac{1}{x+a}}x^{\frac{x}{x+a}}a^{\frac{a}{a+x}}$$

And :

$$g(x)=\left(ax\right)^{\frac{1}{x+a}}x^{\frac{x}{x+a}}a^{\frac{a}{a+x}}$$

And finally :

$$g''(x)=h(x)$$

Then show that :

$$h(e)<f(e)$$

Some similar attempt :

About an inequality wich is an upper bound for Am-Gm. here we can find some works around an inequality I try to show during less than a week .

The inequality above seems very curious and I cannot explain how I find it (expect using Desmos and $e=2.718...$)

How to show it ?How to explain it ?

Thanks !

One more nice problem ! – Claude Leibovici Dec 09 '21 at 15:40 — Claude Leibovici, Dec 09 '21 at 15:40
@ClaudeLeibovici Thanks you at yeast (hefe in German)^^. – Miss and Mister cassoulet char Dec 09 '21 at 15:56 — Miss and Mister cassoulet char, Dec 09 '21 at 15:56

score 1 · Answer 1 · answered Dec 10 '21 at 03:43

Hoping no mistake, for $x=e$ and $a=\frac 12$, we have $$f(e)=\frac{\left(\frac{1}{2}+e\right) \left(\left(\frac{1}{2}+e\right)^2 \left(\frac{15}{2}+e\right)-4\right)}{\left(\frac{5}{2}+e\right)^3}-8^{\frac{1}{-1- 2 e}} e^{1+\frac{1}{1+2 e}}$$ $$h(e)=\frac{2^{1-\frac{3}{1+2 e}} e^{\frac{1}{1+2 e}-1} \left(1+e \left(-1+2 e \left(-3-2 e+9 \log ^2(2)+6 \log (2)\right)+12 \log (2)\right)\right)}{(1+2 e)^4}$$

Computed exactly $$f(e)=0.007813852236\qquad \text{and} \qquad h(e)=0.007821413387$$ $$f(e)-h(e)=-7.56115 \times 10^{-6}$$

Then $h(e) > f(e)$

Show that $h(e)

Problem :

$$g(x)=\left(ax\right)^{\frac{1}{x+a}}x^{\frac{x}{x+a}}a^{\frac{a}{a+x}}$$

Some similar attempt :

1 Answers1