Prove that if $a_k \in [0, 1] $ then $\frac {1}{1+a_1} + \frac {1}{1+a_2} + \cdots +\frac {1}{1+a_n} \le \frac{n}{1+ \sqrt[n]{a_1a_2\cdots a_n}}$.

Question

One of my friends gave me this problem:

If $a_k \in [0, 1]$, Prove that: $$\frac {1}{1+a_1} + \frac {1}{1+a_2} + \cdots +\frac {1}{1+a_n} \le \frac{n}{1+ \sqrt[n]{a_1a_2\cdots a_n}}$$

I have been trying this all day trying to apply things from AM-GM to Cauchy to Chebyshev to Convex functions. Nothing worked.

Can anyone give a solution or help. Much thanks.

You shouldn't feel bad for not solving it. Most because it's false. $n=2, a_1 = 1, a_2 = 0$. — mathworker21, Nov 29 '18 at 18:03
Im sorry I made a mistake but the RHS is less than. I edited the question. Sorry. — ujwal kumar, Nov 29 '18 at 18:08
Essentially a duplicate of Prove that $\frac{1}{1+x_1}+\frac{1}{1+x_2}+\cdots+\frac{1}{1+x_n} \geq \frac{n}{\sqrt[n]{x_1x_2\cdots x_n}+1}$ – via the substitution $x_i = 1/a_i$ — Martin R, Nov 29 '18 at 19:09

score 6 · Accepted Answer · answered Nov 29 '18 at 18:52

Suppose $a_i = e^{\lambda_{i}}$ for all $i$. Define $f$ : $x \mapsto \frac{1}{1+e^x}$. $f$ is concave on $]-\infty, 0]$, so $\frac{1}{1+e^{\frac{1}{n}\sum_{i=1}^n \lambda_i}} \geq \frac{1}{n}\sum_{i=1}^{n}\frac{1}{1+e^{\lambda_{i}}}$, which allows us to conclude.

Prove that if $a_k \in [0, 1] $ then $\frac {1}{1+a_1} + \frac {1}{1+a_2} + \cdots +\frac {1}{1+a_n} \le \frac{n}{1+ \sqrt[n]{a_1a_2\cdots a_n}}$.

1 Answers1