Inequality involving AM - GM: $\sum_{i=1}^n \frac{1}{1+a_i} \ge \frac{n}{1+{{(a_1.a_2...a_n)}}^{1/n}}$

Question

given that $a_i > 1$, how do I prove that $$\sum_{i=1}^n \frac{1}{1+a_i} ≥ \frac{n}{1+{{(a_1.a_2...a_n)}}^{1/n}}$$ by applying AM - GM inequality? Thanks in advance!

Answer 1

I don't know about AM/GM, but you can get it by applying Jensen's inequality to $$f(x)=\frac1{1+\exp(x)}$$ which is a convex function for $x>0$. Then $$\sum_{i=1}^n f(b_i)\ge nf\left(\frac1n\sum_{j=i}^n b_i\right)$$ for $b_i>0$. Set $b_i=\ln a_i$.

Answer 2

According to AM/GM inequality,

$$ \frac{1}{n}\sum_{i=1}^{n} \frac{1}{1+a_i} \ge \prod_{i=1}^{n} \left( \frac{1}{1+a_i} \right)^{\frac{1}{n}}. $$

Now let's see the RHS expression.

$$ (1+a_1)(1+a_2) > 1+a_1a_2 \qquad \text{(as $a_i > 0$)} $$

Similarly,

$$ \prod_{i=1}^{n} (1+a_i) > 1 + P_n \tag{1} $$

and you can also proof that $\prod_{i=1}^{n} (1+a_i) < \frac{1}{1-P_n}$. Since the reciprocal of the expression $\text{(1)}$ can be seen as

$$ \frac{1}{n}\sum_{i=1}^{n} \frac{1}{1+a_i} \ge \prod_{i=1}^{n} \left(\frac{1}{1+a_i}\right)^{\frac{1}{n}} > \left(\frac{1}{1+P_n}\right)^{\frac{1}{n}} > \frac{1}{1+P_n^{1/n}} $$

Hope this helps.

Answer 3

Go by the definition of AM-GM.We see-

$$\frac{\frac{1}{1+a_1}+\frac{1}{1+a_2}+...+\frac{1}{1+a_n}}{n}\geq\sqrt[n]\frac{1}{(1+a_1)(1+a_2)...(1+a_n)}$$

$$\implies\frac{1}{1+a_1}+\frac{1}{1+a_2}+...+\frac{1}{1+a_n}\geq\frac{n}{\{(1+a_1)(1+a_2)...(1+a_n)\}^\frac{1}{n}}$$

Now,$(1+a_1)(1+a_2)=1+a_1a_2+a_1+a_2\geq1+a_1a_2$.So,by induction,you can show that $(1+a_1)(1+a_2)...(1+a_n)\geq1+(a_1a_2...a_n)$.

From here it follows that,$$\frac{1}{1+a_1}+\frac{1}{1+a_2}+...\frac{1}{1+a_n}\geq\frac{n}{\{1+(a_1a_2...a_n)\}^\frac1n}\geq\frac{n}{1+(a_1a_2...a_n)^\frac1n}$$