Home » Uncategorized » Inequality 35(Gabriel Dospinescu)

# Inequality 35(Gabriel Dospinescu)

Problem:

If $\displaystyle a_1,a_2,...,a_n$ are positive real numbers, such that $\displaystyle a_{1}\cdot a_2\cdot...\cdot a_n=1$, prove that

$\displaystyle \sum^{n}_{i=1}a_i\geq\sum^{n}_{i=1}\sqrt{\frac{a^{2}_{i}+1}{2}}$.

Solution:

From the AM-GM we deduce that

$\displaystyle a_{i}+1+\frac{2(a^{2}_{i}+1)}{a_i+1}\geq 2\sqrt{2}\sqrt{a^{2}_{i}+1}$.

So,

$\displaystyle \sum^{n}_{i=1}(a_i+1)+2\sum^{n}_{i=1}\frac{a^{2}_{i}+1}{a_i+1}\geq 2\sqrt{2}\sum^{n}_{i=1}\sqrt{a^{2}_{i}+1}$.

Divide the last relation by $\displaystyle 4$. Then we get that

$\displaystyle \frac{1}{4}\sum^{n}_{i=1}(a_i+1)+\frac{1}{2}\sum^{n}_{i=1}\frac{a^{2}_{i}+1}{a_i+1}\geq \frac{1}{\sqrt{2}}\sum^{n}_{i=1}\sqrt{a^{2}_{i}+1}$,

or

$\displaystyle \frac{1}{4}\sum^{n}_{i=1}(a_i+1)+\frac{1}{2}\sum^{n}_{i=1}\frac{a^{2}_{i}+1}{a_i+1}\geq \sum^{n}_{i=1}\sqrt{\frac{a^{2}_{i}+1}{2}}$.

So, it is enough to prove that

$\displaystyle \frac{1}{4}\sum^{n}_{i=1}(a_i+1)+\frac{1}{2}\sum^{n}_{i=1}\frac{a^{2}_{i}+1}{a_i+1}\leq \sum^{n}_{i=1}a_i\Longrightarrow 3\sum^{n}_{i=1}a_i\geq 2\sum^{n}_{i=1}\frac{a^{2}_{i}+1}{a_i+1}+n$.

But it holds that,

$\displaystyle \sum^{n}_{i=1}\frac{a^{2}_{i}+1}{a_i+1}=\sum^{n}_{i=1}(a_i+1)-\sum^{n}_{i=1}\frac{2a_i}{a_i+1}$.

From this we need to prove now

$\displaystyle 3\sum^{n}_{i=1}a_i\geq 2\sum^{n}_{i=1}(a_i+1)-4\sum^{n}_{i=1}\frac{a_i}{a_i+1}+n$.

That is,

$\displaystyle \sum^{n}_{i=1}(a_i+1)+4\sum^{n}_{i=1}\frac{a_i}{a_i+1}\geq 4n$,

which is obvious from the AM-GM inequality, Q.E.D.