# Typical Olympiad Inequality? If $\sum_i^na_i=n$ with $a_i>0$, then $\sum_{i=1}^n\left(\frac{a_i^3+1}{a_i^2+1}\right)^4\geq n$

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 $$b_i=x^{\frac{1}{2}}$$

• 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 at 18:34
• 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 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 at 20:41
• @MichaelRozenberg: How can you know what OP attempted to solve the problem? – Martin R Oct 13 at 19:16
• @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 at 20:21

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 $$\left(\frac{x+1}{2}\right)^4\geq\left(\sqrt{x}\right)^4=x^2.$$

• Please pardon my ignorance , but could you include more details on your use of rearrangement ? – Aspirant Oct 13 at 12:50
• @Aspirant I added something. See now. – Michael Rozenberg Oct 13 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 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 at 14:06
• @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 at 14:37

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 $$a_1=a_2=\ldots=a_n=1$$.

• 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 at 0:39
• @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 at 1:11
• @Aspirant There is a simple way. Show your attempts and I'll show this way. – Michael Rozenberg Oct 13 at 5:18
• @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 at 6:21
• 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 at 7:04