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# Inequality 47(Christos Patilas)

Problem:

If $\displaystyle x_i$ for $\displaystyle i=1,2,...,n$ are positive real numbers then prove that

$\displaystyle \sum^{n}_{i=1}\left(5\sqrt[5]{x^{3}_{i}}-3\sqrt[3]{\left(\frac{3x_i+2}{5}\right)^5}\right)\leq 2n$

Solution(An Idea by Vo Quoc Ba Can):

We only need to prove that

$\displaystyle 5\sqrt[5]{a^3}-3\sqrt[3]{\left(\frac{3a+2}{5}\right)^5}\leq 2$ for all $\displaystyle a>0$.

So, using the AM-GM Inequality we have that

$\displaystyle a+a+a+1+1\geq 5\sqrt[5]{a^3}$.

It follows that

$\displaystyle \sqrt[3]{\left(\frac{3a+2}{5}\right)^5}\geq \sqrt[3]{\left(\sqrt[5]{a^3}\right)^5}=a$.

Therefore it suffices to prove that

$\displaystyle 5\sqrt[5]{a^3}-2\leq 3a$,

which is obviously true from the AM-GM Inequality, Q.E.D.

## 6 thoughts on “Inequality 47(Christos Patilas)”

1. chao ban 😀 ban hay giup minh chung minh mot bat dang thuc sau day:
cho cac so duong a , b , c thoa man $a\leq b\leq c$ chung minh rang neu ham
f la mot ham thoa man f”>0 va f(0)=0 thi ta nhan duoc bat dang thuc sau day:
$\displaystyle f(c)\geq \frac{c}{b}f(b-a)$

• Sorry my friend but i do not know Vietnamese, thus i cannot respond to your question.

2. Hello Mr !!! Now i want you prove the solution of the inequality below:
Let positive numbers (a , b) and (p, q) ; $\displaystyle \frac{1}{p}+\frac{1}{q}=1$ .
Prove that:
$\displaystyle \frac{1}{p}a^p+\frac{1}{q}b^q\geq 2\biggl(\frac{1}{p}\biggl)^{\frac{1}{p}}.\biggl(\frac{1}{q}\biggl)^{\frac{1}{q}}ab$

3. Aw, this was an extremely nice post. Finding the time
and actual effort to make a really good article… but what can I say…
I procrastinate a lot and never manage to get anything done.