# Show that $\frac{2a_1^2}{a_1+a_2}+\frac{2a_2^2}{a_2+a_3}+...+\frac{2a_n^2}{a_n+a_1}\geq a_1+a_2+...+a_n$ [duplicate]

Showing that $\frac{2a_1^2}{a_1+a_2}+\frac{2a_2^2}{a_2+a_3}+...+\frac{2a_n^2}{a_n+a_1}\geq a_1+a_2+...+a_n$ holds for positive $a_i$s.

I've tried adding $a_1+a_2, a_2+a_3,...,a_n+a_1$ respectively to the terms of the LHS and using AM-GM on them, but it didn't get me to the needed point. Also tried AM GM on the whole LHS, but couldn't get much. How is it done?

• Oct 16, 2019 at 13:59

You could also use the Cauchy-Schwarz Inequality (http://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality). Note that $$\left(\frac{a_1+a_2}{2}+\frac{a_2+a_3}{2}+...+\frac{a_n+a_1}{2}\right)\left(\frac{2a_1^2}{a_1+a_2}+...+\frac{2a_n^2}{a_n+a_1}\right)\ge (a_1+a_2+...+a_n)^2.$$Dividing both sides by $a_1+a_2+...+a_n$ gives the desired conclusion.

• Very simple and elegant solution. Nov 23, 2014 at 0:33

Consider the function $f(x)=\frac{2}{1+x}$ for $x>0$. Then $f$ is convex so with $S=\sum_{i=1}^na_i$, we can use Jensen's inequality to infer that $$\frac{a_1}{S}f(a_2/a_1)+\cdots+\frac{a_n}{S}f(a_1/a_n)\geq f\left(\frac{a_2}{S}+\frac{a_3}{S}+\cdots+\frac{a_1}{S}\right)=f(1)=1.$$ Thus, $$\sum_{i=1}^na_i=S\leq a_1f(a_2/a_1)+\cdots+a_nf(a_1/a_n)=\frac{2a_1}{1+a_2/a_1}+\cdots+\frac{2a_n}{1+a_1/a_n}$$ which simplifies to the LHS of your inequality.

Here is another way: $$\frac{2a_k^2}{a_k+a_{k+1}} + \frac{a_k + a_{k+1}}2 \ge 2a_k$$ by AM-GM. Sum this cyclically and simplify to get your inequality.

$$\frac{2a_1^2}{a_1+a_2}+\frac{2a_2^2}{a_2+a_3}+...+\frac{2a_n^2}{a_n+a_1}\geq a_1+a_2+...+a_n\tag{1}$$

Thus to prove (1) hold, it is suffice for us to add $2A(a_1+a_2+...+a_n)$ on both sides and and to prove:

$$\frac{2a_1^2}{a_1+a_2}+A(a_1+a_2)+\frac{2a_2^2}{a_2+a_3}+A(a_2+a_3)+$$ $$+...+\frac{2a_n^2}{a_n+a_1}+A(a_n+a_1)\geq (2A+1)(a_1+a_2+...+a_n).\tag{2}$$

Since $\frac{2a_1^2}{a_1+a_2}+A(a_1+a_2)\geq 2\sqrt{2A}a_1$, we have to prove:

$$2\sqrt{2A}(a_1+a_2+...+a_n)\geq (2A+1)(a_1+a_2+...+a_n)\tag{3}$$ or $$2\sqrt{2A}\geq 2A+1\tag{4}$$

And we found out that $A=1/2$ is the only solution.

• Cool, but could you please explain why is A=1/2 a solution? $\sqrt{2}$ is less than 1+1=2. I just don't get it, does it mean that we have to add 1/2(a_1+a_2), etc., to each of the LHS terms? Nov 22, 2014 at 9:45
• For $A=1/2$, we have $2\sqrt{2*1/2}=2=2*(1/2)+1$. So $A=1/2$ is a solution to (4).
– mike
Nov 22, 2014 at 9:48
• @mike:Why should $(3)$ hold? Nov 22, 2014 at 10:43
• I think what he means is that if (3) can somehow hold true, then the conclusion would follow. In order for (3) to be true, (4) must be true, and $A=\frac{1}{2}$ is a good choice for (4) to hold true. This is like working backwards to find a good value of $A$ to get the conclusion. Nov 22, 2014 at 15:25
• Moreover, I would like to emphasize that what is shown here is a technique to find the value of A. It is better to determine the value of A later to suit the purposes of the problem. When writing up the proof however, the part about the values of A would not be included, but the motivation behind why a certain value was chosen would be lost. Nov 22, 2014 at 19:53