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Inequality 47(Christos Patilas)


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)

  2. 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.

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