11

Let $\sum_i^na_i=n$, $a_i>0$. Then prove that $$ \sum_{i=1}^n\left(\frac{a_i^3+1}{a_i^2+1}\right)^4\geq n $$

I have tried AM-GM, Cauchy-Schwarz, Rearrangement etc. but nothing seems to work. The fourth power in the LHS really evades me, and I struggle to see what can be done.

My attempts didn’t lead me to any result ... Simply cauchy , where $a_i=x,$ $b_i=1$ to find an inequality involving $\sum x^2$ . I also tried finding an inequality involving $\sum x^3$ using $a_i=\frac{x^3}{2}$ and $b_i=x^{\frac{1}{2}}$

Michael Rozenberg
  • 203,855
Aspirant
  • 417
  • 1
    I solved your problem. If you want to see my solution, show please how exactly you tried AM-GM, C-S and Rearrangement. – Michael Rozenberg Oct 12 '19 at 18:34
  • 1
    If we let $f(x)=\left(\frac{x^3+1}{x^2+1}\right)^4$, then I think that it would not be too hard to prove that $$f(x)\geq f'(1)(x-1) + f(1)$$ for all $x\geq 0$. If you can prove that, then the rest seems to be rather straight forward. Am I missing something? – irchans Oct 12 '19 at 18:59
  • @irchans You are very right. It is not difficult to prove that $f(x)\geq 2x-1$ for all $x\geq 0$. – Batominovski Oct 12 '19 at 20:41
  • @MichaelRozenberg: How can you know what OP attempted to solve the problem? – Martin R Oct 13 '19 at 19:16
  • 1
    @MartinR In this case, from a comment by the OP, I think it's legitimate to add such. (Whether it's useful is a different kettle of fish.) – Daniel Fischer Oct 13 '19 at 20:21

2 Answers2

7

You need to use another queue:

By Rearrangement, AM-GM and C-S we obtain: $$\sum_{i=1}^n\left(\frac{a_i^3+1}{a_i^2+1}\right)^4\geq \ \sum_{i=1}^n\left(\frac{a_i+1}{2}\right)^4\geq\sum_{i=1}^na_i^2=\frac{1}{n}\sum_{i=1}^n1^2\sum_{i=1}^na_i^2\geq\frac{1}{n}\left(\sum_{i=1}^na_i\right)^2=n. $$

I used the following. $$\frac{x^3+1}{x^2+1}\geq\frac{x+1}{2}$$ it's $$2(x^3+1)\geq(x^2+1)(x+1)$$ or $$x^3+1\geq x^2+x$$ or $$x^2\cdot x+1\cdot1\geq x^2\cdot1+x\cdot1,$$ which is true by Rearrangement because $(x^2,1)$ and $(x,1)$ have the same ordering.

Also, by AM-GM $$\left(\frac{x+1}{2}\right)^4\geq\left(\sqrt{x}\right)^4=x^2.$$

Michael Rozenberg
  • 203,855
  • Please pardon my ignorance , but could you include more details on your use of rearrangement ? – Aspirant Oct 13 '19 at 12:50
  • @Aspirant I added something. See now. – Michael Rozenberg Oct 13 '19 at 12:55
  • Ty for your solution . As is quite apparent , I’m terrible at inequalities . It would greatly help me if you describe how you approached this problem , the first thing you did , etc. I would also greatly appreciate tips for inequality problems in general . – Aspirant Oct 13 '19 at 13:17
  • @Aspirant I saw very many ways to the proof and chose something simple. Your second question is very very hard. For example, can you explain, if you need to say something, why you choose just words, which you said and not something other? Usually you pronounce a lot of words! Just prove inequalities as often as you say, and you'll know. – Michael Rozenberg Oct 13 '19 at 14:06
  • I have accepted this solution , but Batominovski mentions that you have another solution , that is even more elegant. I would greatly appreciate it if you posted that as well... – Aspirant Oct 13 '19 at 14:26
  • 2
    @Aspirant I posted my solution $150$ minutes before the Batominovski's solution. I just hided my solution from you. He meant the solution, which you see now. By the way, the method, which Batominovski showed and irchans said in his comment, named Tangent Line method. Also, we can use Rearrangement in the inequality $\frac{x^3+1}{x^2+1}\geq\sqrt{x}$, which gives a solution with one step less. – Michael Rozenberg Oct 13 '19 at 14:37
  • +1. Engineous. How did you see the first inequality in the first place, given that the proof took two steps? – Hans Oct 13 '19 at 21:05
  • @Hans It's hard to explain. Maybe I just saw before similar inequalities... – Michael Rozenberg Oct 13 '19 at 21:52
4

A Hint for an Alternative Solution.

We want to show that $$\left(\frac{x^3+1}{x^2+1}\right)^4\geq 2x-1$$ for every $x\in\mathbb{R}$. By the AM-GM Inequality, $$\left(\frac{x^3+1}{x^2+1}\right)^4+1\geq 2\,\left(\frac{x^3+1}{x^2+1}\right)^2\,.$$ Hence, it suffices to verify that $$\left(\frac{x^3+1}{x^2+1}\right)^2\geq x$$ for all $x\in\mathbb{R}$. This part is left for the OP.

Remark. From this solution, we need not require that $a_1,a_2,\ldots,a_n$ be positive. That is, for any real numbers $a_1,a_2,\ldots,a_n$ such that $\sum\limits_{i=1}^n\,a_i=n$, we always have $$\sum_{i=1}^n\,\left(\frac{a_i^3+1}{a_i^2+1}\right)^4\geq n\,.$$ However, the sole equality case is when $a_1=a_2=\ldots=a_n=1$.

Batominovski
  • 50,341
  • It's amazing how often simple inequalities like the AM-GM inequality are used. I would have bounded derivatives and maybe used Taylor series which would have been much more work. – irchans Oct 13 '19 at 00:39
  • @Batominovski I was able to prove that (x^3+1)/(x^2+1) is >= x by analysing the functions . There is equality at x=1 , and the local minima of the LHS also happens to be at x=1. However , is there any way of using a simple inequality to prove it? – Aspirant Oct 13 '19 at 01:11
  • @Aspirant There is a simple way. Show your attempts and I'll show this way. – Michael Rozenberg Oct 13 '19 at 05:18
  • @Aspirant You can apply AM-GM once more time to get $$\left(\frac{x^3+1}{x^2+1}\right)^2+1\geq 2,\left|\frac{x^3+1}{x^2+1}\right|,.$$ Now it suffices to prove that $$\left|\frac{x^3+1}{x^2+1}\right|\geq \frac{x+1}{2}$$ for every $x\in\mathbb{R}$. From here on, there are multiple ways to do it. – Batominovski Oct 13 '19 at 06:21
  • I think it is ok to leave hints, if that is what you are concerned. I think your solution is great, though. I didn't fully reveal my whole proof and leave bits and pieces for the OP to figure out. – Batominovski Oct 13 '19 at 07:04
  • @MichaelRozenberg But my attempts didn’t lead me to any result ... Simply cauchy , where $a_i=x$, $b_i=1$ to find an inequality involving $\sum x^2$ . I also tried finding an inequality involving $\sum x^3$ using $a_i=x^{3/2}$ and $b_i = x^{1/2}$ – Aspirant Oct 13 '19 at 08:58
  • You should get something though. For $x\geq 0$, by the Cauchy-Schwarz Inequality, $$(x^3+1)(x+1)\geq (x^2+1)^2$$ so $$\frac{x^3+1}{x^2+1}\geq \frac{x^2+1}{x+1},.$$ Then, use the Cauchy-Schwarz Inequality again to show that $$\frac{x^2+1}{x+1}\geq \frac{x+1}{2}$$ for $x\geq 0$. But this is also the approach used by Michael Rozenberg in his hidden solution (but he has a different solution, a very nice one). But as I said, there are multiple ways to show that $\dfrac{x^3+1}{x^2+1}\geq \dfrac{x+1}{2}$ (for $x\geq 0$). I know at least three more. – Batominovski Oct 13 '19 at 09:02
  • Even the inequality $\left(\dfrac{x^3+1}{x^2+1}\right)^2\geq x$ for $x\geq 0$ has at least two more proofs. – Batominovski Oct 13 '19 at 09:11
  • @Batominovski We can use that $$(x^3+1)^2-x\cdot(x^2+1)^2=(x-1)^2\cdot(x^4+x^3+x^2+x+1)$$. What were the two proofs you thought of? – Maximilian Janisch Oct 15 '19 at 07:22
  • That's one proof I had. The other is this: for $x\geq 0$, $$\begin{align}(x^3+1)^2&=(x+1)^2\left(x^2-x+1\right)^2\&\geq (2\sqrt{x})^2,\left(\frac{x^2+1}{2}+\frac{(x-1)^2}{2}\right)^2\&\geq (2\sqrt{x})^2,\left(\frac{x^2+1}{2}\right)^2 =x,\left(x^2+1\right)^2,.\end{align}$$ – Batominovski Oct 15 '19 at 07:32