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it's related to this Proving $\ln \cosh x\leq \frac{x^2}{2}$ for $x\in\mathbb{R}$

The case $x\geq 2$ is easy so continue with :

The case $0\leq x \leq 2$

We start with a proof of the Young inequality see here

Purely "algebraic" proof of Young's Inequality

If we put $a=e^{2x}$$ \quad $$b=e^{-x}$$ \quad $$p=\frac{x}{4}$ we get: $$\frac{4e^{0.5x^2}}{x}+\frac{e^{\frac{-x^2}{x-4}}}{\frac{x}{x-4}}\geq e^x$$ Or : $$e^{0.5x^2}\geq (e^x-e^{\frac{-x^2}{x-4}}\frac{x-4}{x})\frac{x}{4}$$ Or : $$e^{0.5x^2}\geq e^x(\frac{x}{4}) -(e^{\frac{-x^2}{x-4}})\frac{x-4}{x}\frac{x}{4}$$

Futhermore it's easy to remark that we have :

$e^x\geq \frac{x}{4}$ et $\quad$$e^{\frac{-x^2}{x-4}}\geq \frac{4-x}{4}$

Wich follow from the well-know inequality : $e^x\geq x+1$

Now we apply the Tchebychev's inequality for 4 variables : $$e^x(\frac{x}{4}) -(e^{\frac{-x^2}{x-4}})\frac{x-4}{x}\frac{x}{4}\geq 0.5(e^x+e^{\frac{-x^2}{x-4}})(\frac{x}{4}+\frac{4-x}{4})$$ And we have : $$0.5(e^x+e^{\frac{-x^2}{x-4}})\geq 0.5(e^x+e^{-x})$$ For the rest it is enough to think about symmetry

So we get a pure algebraic proof .

My question is :Can you simplify this or use an other way ? And if I'm wrong correct me

Thanks a lot

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  • $\begingroup$ You should tag this with something more advanced than just algebra-precalculus. That'll get your question more attention. $\endgroup$ – Mike Pierce Nov 5 '17 at 2:38
  • $\begingroup$ What do you mean by a "pure algebraic proof" of an inequality concerning the function $\log\circ\cosh$? $\endgroup$ – Christian Blatter Nov 26 '17 at 16:46
  • $\begingroup$ Yes it's ambiguous .I mean without tools like the derivative by example. $\endgroup$ – user448747 Nov 26 '17 at 17:21
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A Chinese proverb (more precisely, a quotation from Chapter 64 of “Tao Te Ching” ascribed to Lao Tse) says that a journey of a thousand miles begins with a single step. Unfortunately, you made an error already at the start. If $0<x\le 2$ then $p=\frac x4<1$, so $\frac 1q=1-\frac 1p<0$, thus $q<0$ and you cannot apply Young’s inequality.

A straightforward “algebraic” way is to listen to a hint of the Taylor series an use a more precise bound $e^x\ge 1+x+\frac{x^2}{2}+\frac{x^3}{6}$ for $x\ge 0$. This inequality should be proved the same way as a used in your proof well-known inequality $e^x\ge 1+x$. (By the way, how to prove it “without tools like the derivative”?) But if you still insist, I can remark that for each natural $n\ge 3$ we have

$$e^x=e^{\frac xn\cdot n}\ge \left(1+\frac xn\right)^n\ge$$ $$1+n\cdot \frac xn+{n\choose 2}\left(\frac xn\right)^2+{n\choose 3}\left(\frac xn\right)^3=$$ $$1+x+\frac{n-1}{n}\cdot \frac 12\cdot x^2+\frac{(n-1)(n-2)}{n^2}\cdot \frac 16\cdot x^3.$$

Lets turn back to the initial inequality. Changing the sign of $x$, if needed, we may assume $x\ge 0$. Dividing both sides of the inequality by $e^x$, we see that it suffices to prove that

$$\frac {1+e^{-2x}}2\le e^{\frac {x^2}2-x}.$$

But $$e^{x^2-x}\ge 1+ x^2-x+\frac 12\left(\frac{x^2}2-x\right)^2=1-x+x^2-\frac 12x^3+\frac 18x^4.$$

On the other hand,

$$e^{-2x}=\frac 1{e^{2x}}\le \frac{1}{1+2x+2x^2+\frac 43x^3}.$$

So it suffices to check that

$$4+\frac{12}{3+6x+6x^2+4x^3}\le 8-8x+8x^2-4x^3+x^4$$

$$\frac{12}{3+6x+6x^2+4x^3}\le 4-8x+8x^2-4x^3+x^4$$

$$12\le (4-8x+8x^2-4x^3+x^4)(3+6x+6x^2+4x^3)$$

$$12\le 12+4x^3-5x^4+14x^5-10x^6+4x^7$$

$$0\le 4-5x+14x^2-10x^3+4x^4$$

The last inequality holds because $4+2x^2\ge 4\sqrt{2}x\ge 5x$ and $12x^2+4x^4\ge 8\sqrt{3}x^3\ge 10x^3$ by AM-GM.

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