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# Inequality 10(Christos Patilas)

Problem:

If $\displaystyle a,b,c\in \mathbb{R}$, prove that

$\displaystyle \sqrt[4]{a^4+b^4+c^4+1}\geq \sqrt[5]{a^5+b^5+c^5+1}$.

Solution:

We have that

$\displaystyle \sqrt[5]{\sum_{cyc}a^5+1}\leq \sqrt[5]{\sum_{cyc}\left|a\right|^5+1}$,

so we only need to prove the inequality for $\displaystyle a,b,c\geq 0$.

Let us write the given inequality into the form

$\displaystyle \left(\sum_{cyc}a^4+1\right)^{\frac{5}{4}}\geq \sum_{cyc}a^5+1$ and divide with $\displaystyle \left(\sum_{cyc}a^4+1\right)^{\frac{5}{4}}$.

Then we acquire

$\displaystyle \sum_{cyc}\frac{a^5}{\left(\sum_{cyc}a^4+1\right)^{\frac{5}{4}}}+\frac{1}{\left(\sum_{cyc}a^4+1\right)^{\frac{5}{4}}}\leq 1$

or

$\displaystyle \sum_{cyc}\left(\frac{a^4}{\sum_{cyc}a^4+1}\right)^{\frac{5}{4}}+\frac{1}{\left(\sum_{cyc}a^4+1\right)^{\frac{5}{4}}}\leq 1$.

But for all non-negative numbers holds that

$\displaystyle \left(\frac{a^4}{\sum_{cyc}a^4+1}\right)^{\frac{5}{4}}\leq \frac{a^4}{\sum_{cyc}a^4+1}$.

So from the above inequality we get that

$\displaystyle \sum_{cyc}\left(\frac{a^4}{\sum_{cyc}a^4+1}\right)^{\frac{5}{4}}+\frac{1}{\left(\sum_{cyc}a^4+1\right)^{\frac{5}{4}}}\leq \frac{\sum_{cyc}a^4+1}{\sum_{cyc}a^4+1}=1$, Q.E.D.