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For $0<t\leq 1$ show $$ \text{ln}(t)\geq \frac{t-1}{2t+2}\left(1+\sqrt{\frac{2t^2+5t+2}{t}}\right)$$

On a side note, I've been taking this summer class on inequalities. The professor told us that his main source are all sorts of math olympiads. Given that I'm gonna take Calc 3 next semester and Real Analysis afterwards, are those types of inequalities really as essential as my prof suggests? As you can tell, I'm not really capable of proving most of them by myself yet...

InanimateBeing
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Octot
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  • Have you tried simple thing like $x=(0.5y-2)$? or use exponential ? – Miss and Mister cassoulet char Sep 03 '22 at 10:26
  • Can't really make anything of you substitution. I have tried using expponential but it didn't lead me to anything – Octot Sep 03 '22 at 11:09
  • Let $0<x\leq 1 $ and $n$ a positive natural number then define $f\left(x\right)=\left(x^{2}-1\right)\ln\left(x^{n}\right)-2nx^{2}$ the second derivative is positive so using a tangent line we have better result with $$\frac{\left(f'\left(1\right)\left(x^{\frac{1}{n}}-1\right)+f\left(1\right)+2nx^{\frac{2}{n}}\right)}{\left(x^{\frac{2}{n}}-1\right)}$$ more than $n(x^{1/n}-1)$ – Miss and Mister cassoulet char Sep 03 '22 at 11:10
  • Then we can make the substitution $y=1/x$ and see what's happend ... – Miss and Mister cassoulet char Sep 03 '22 at 11:57
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    In real analysis you will need to do several estimates while proving things. They are very rearly as tight as the contest questions on the theme. But some themes (AM-GM, Young, Jensen, Hölder) are very important. – Jyrki Lahtonen Sep 04 '22 at 20:21
  • @Octot Hey, I wrote an answer some days ago but there is no response. Please see if it is correct and helps you with the basic methodology. Thanks! – An_Elephant Sep 13 '22 at 13:20

3 Answers3

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My try .

here (Which is greater $\frac{13}{32}$ or $\ln \left(\frac{3}{2}\right)$) Michael Rozenberg show :

let $0<x\leq 1$ :

$$\ln{x}\geq(x-1)\sqrt[3]{\frac{2}{x^2+x}}$$

Now we need to show :

$$(x-1)\sqrt[3]{\frac{2}{x^2+x}}\geq \frac{x-1}{2x+2}\left(1+\sqrt{\frac{2x^{2}+5x+2}{x}}\right)$$

or :

$$\sqrt[3]{\frac{2}{x^2+x}}\geq \frac{1}{2x+2}\left(1+\sqrt{\frac{2x^{2}+5x+2}{x}}\right)$$

I'm stuck here .Hope someone can achieve this .

Edit :

In fact using Wolfram Alpha we can conclude quicly see this logarithmic derivative link

Just some remarks for another proof :

We have the inequality :

$$\sqrt{\frac{\left(2x^{2}+5x+2\right)}{x}}\geq \frac{3}{2}\left(\sqrt{\frac{1}{x^{a}}}+\sqrt{x^{a}}\right)$$

Where :

$$a=\frac{2\sqrt{2}}{3}$$

So we need to show :

$$\left(\frac{3}{2}\left(\sqrt{\frac{1}{x^{a}}}+\sqrt{x^{a}}\right)+1\right)\left(x-1\right)\cdot\frac{1}{\left(2x+2\right)}-\ln\left(x\right)\leq 0$$

Or $x\to x^{\frac{2}{a}}$ :

$$\left(\frac{3}{2}\left(\frac{1}{x}+x\right)+1\right)\left(x^{\frac{2}{a}}-1\right)\cdot\frac{1}{\left(2x^{\frac{2}{a}}+2\right)}-\frac{\ln\left(x\right)2}{a}\leq 0$$

Now we substitute $\ln(x)=y$ and then we can use derivative .

Miss and Mister cassoulet char
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0

Consider only strict inequality by dropping off the equal sign (because equality can be proven easily by putting t = 1).

Let $A$ = ${\frac{(t-1)}{2(t+2)}}$= $(1/2)- 1/(t+1)$

Now, t∈ (0,1) . Hence,

$A = (1/2)- 1/(t+1)∈ (-1/2,0)$

Let $B$ = $\left(1+\sqrt{\frac{2t^2+5t+2}{t}}\right)$

Now, we know that ;

$t + (1/t) >2$

$2[t + (1/t)] >4$

$2[t + (1/t)] + 5 >9$

$\sqrt{2[t + (1/t)] + 5} >3$

$\sqrt{2[t + (1/t)] + 5} + 1 >4$

Hence,

$B$ $> 4$ because

$\sqrt{2[t + (1/t)] + 5} + 1 = \left(1+\sqrt{\frac{2t^2+5t+2}{t}}\right)$

Hence, your inequality reduces to ;

$ln(t) > AB$

where $A∈ (-1/2,0)$ and $B ∈ (4,∞)$

Can you proceed further?

An_Elephant
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Fact 1: If $x \in (0, 1]$, then $$\ln x \ge \frac{(x - 1)(1 + x^{1/3})}{x + x^{1/3}}.$$ (See [1], page 272. This inequality is due to Karamata.)

By Fact 1, it suffices to prove that $$\frac{(t - 1)(1 + t^{1/3})}{t + t^{1/3}} \ge \frac{t-1}{2t+2}\left(1+\sqrt{\frac{2t^2+5t+2}{t}}\right)$$ or $$\frac{1 + t^{1/3}}{t + t^{1/3}} \le \frac{1}{2t+2}\left(1+\sqrt{\frac{2t^2+5t+2}{t}}\right).$$

Letting $t = u^3$, it suffices to prove that, for all $u\in (0, 1]$, $$\frac{1 + u}{u^3 + u} \le \frac{1}{2u^3+2}\left(1+\sqrt{\frac{2u^6+5u^3+2}{u^3}}\right)$$ or $$\frac{2u^4 + u^3 + u + 2}{u^3 + u} \le \sqrt{\frac{2u^6+5u^3+2}{u^3}}$$ or $$\left(\frac{2u^4 + u^3 + u + 2}{u^3 + u}\right)^2 \le \frac{2u^6+5u^3+2}{u^3}$$ or $$\frac{2(u^2 + u + 1)(u^2 - u + 1)(1 + u)^2(1 - u)^4}{u^3(u^2 + 1)^2} \ge 0$$ which is true.

We are done.

Reference.

[1] D. S. Mitrinovic, P. M. Vasic, “Analytic Inequalities,” 1970.

River Li
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