Proving inequality $(a_{1}+\frac{1}{a_{1}})^{2}+(a_{2}+\frac{1}{a_{2}})^{2}+\cdots+(a_{n}+\frac{1}{a_{n}})^{2} \ge \frac{(n^2+1)^2}{n}$

Question

If $a_n$ is positive real number and $a_{1}+a_{2}+\cdots+a_{n}=1$

prove that :

$$(a_{1}+\frac{1}{a_{1}})^{2}+(a_{2}+\frac{1}{a_{2}})^{2}+\cdots+(a_{n}+\frac{1}{a_{n}})^{2} \ge \frac{(n^2+1)^2}{n}$$

in 6 methods.

First, using cauchy-schwarz inequality and the AM–GM inequality

Second, using cauchy-schwarz inequality and the GM–HM inequality

Third, after expansion using the AM–GM inequality

Fourth, after expansion using Chebyshev's sum inequality

Fifth, given the function $f(x)=(x+\frac{1}{x})^2$, checking the shape of the graph of the function with the second derivative of $f$, and then using jensen's inequality

Sixth, using $sin^2x+cos^2x=1$ and cauchy-schwarz inequality

It's okay to answer few of them, but I'd really appreciate if you answer all of them. Thank you.

Answer 1

We need to prove that $$a_1^2+a_2^2+...+a_n^2+2n+\frac{1}{a_1^2}+\frac{1}{a_2^2}+...+\frac{1}{a_n^2}\geq\frac{(n^2+1)^2}{n}$$ or $$a_1^2+a_2^2+...+a_n^2+\frac{1}{a_1^2}+\frac{1}{a_2^2}+...+\frac{1}{a_n^2}\geq\frac{1}{n}+n^3,$$ which is true by C-S and Holder: $$a_1^2+a_2^2+...+a_n^2=\frac{1}{n}(1+1+...+1)\left(a_1^2+a_2^2+...+a_n^2\right)\geq$$ $$\geq\frac{1}{n}\left(a_1+a_2+...+a_n\right)^2=\frac{1}{n}$$ and $$\frac{1}{a_1^2}+\frac{1}{a_2^2}+...+\frac{1}{a_n^2}=\left(a_1+a_2+...+a_n\right)^2\left(\frac{1}{a_1^2}+\frac{1}{a_2^2}+...+\frac{1}{a_n^2}\right)\geq$$ $$\geq\left(\sqrt[3]{a_1^2\cdot\frac{1}{a_1^2}}+\sqrt[3]{a_2^2\cdot\frac{1}{a_2^2}}+...+\sqrt[3]{a_n^2\cdot\frac{1}{a_n^2}}\right)^3=(1+1+...+1)^3=n^3$$ and we are done!

Answer 2

I will write the proof using Jensen's Inequality.

Let the function $f(x) = \left(x+\dfrac{1}{x}\right)^2 = x^2+2+\dfrac{1}{x^2}$. Then, $f''(x) = 2+\dfrac{6}{x^4} > 0$ for all $x$.

So, since $f$ is convex , for any $a_1,\ldots,a_n$ where $a_1+\cdots+a_n = 1$, we can use Jensen's Inequality to get $$\dfrac{1}{n}\sum_{k = 1}^{n}f(a_k) \ge f\left(\dfrac{1}{n}\sum_{k = 1}^{n}a_k\right)$$ $$\sum_{k = 1}^{n}\left(a_k+\dfrac{1}{a_k}\right)^2 \ge nf\left(\dfrac{1}{n}\right) = n^3+2n+\dfrac{1}{n}.$$ So our proof is done. I leave the remaining cases to you, and I would point out that it may be better to ask for distinct questions in distinct posts.

Answer 3

The shortest proof might be the Cauchy-Schwarz inequality direct application with notice that $1/a_1+1/a_2+\cdots+ 1/a_n \ge n^2$. Can you see this is true?To see the Chebychev inequality, you simply observe that the function $f(x) = x+ 1/x$ is decreasing on $(0,1)$, then WLOG, you can further assume $0 < a_1 \le a_2 \le...\le a_n \implies (f(a_1)^2 \ge (f(a_2))^2 \ge...\ge (f(a_n))^2$. Apply Chebechev inequality standard version $\implies LHS \ge n(a_1(f(a_1))^2+a_2(f(a_2))^2+\cdots + a_n(f(a_n))^2)$, and also Jensen's inequality follows for the $\sum (a_i)^3$, and AM-GM inequality for $\sum 1/a_i$ as above. Can you also see how this works?