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Inequality 49 (Unknown Author)

Problem:

Let \displaystyle a,b,c be positive real numbers such that \displaystyle a+b+c=3, and also let \displaystyle n\geq 12 be a natural number. Prove that

\displaystyle \sum_{cyc}\left(\frac{1}{3a+bc+\frac{n-4}{a}}\right)^{\frac{n}{n-4}}\leq \frac{3}{\sqrt[n-4]{n^n}}.

Solution:

Observe that \displaystyle \frac{1}{3a+bc+\frac{n-4}{a}}=\frac{a}{3a^2+bc+(n-4)}=\frac{a}{a(a+b)(a+c)+(n-4)}. Thus, from the AM – GM Inequality we get that

\displaystyle 4\cdot\frac{a(a+b)(a+c)}{4}+(n-4)\geq n\sqrt[n]{\frac{a^4(a+b)^4(a+c)^4}{4^4}}.

From the above Inequality, we are now capable of building our problem. Specifically, we have that

\displaystyle \frac{a}{a(a+b)(a+c)+(n-4)}\leq \frac{1}{n}\sqrt[n]{\frac{4^4a^{n-4}}{(a+b)^4(a+c)^4}}

or

\displaystyle \left(\frac{a}{a(a+b)(a+c)+(n-4)}\right)^{\frac{n}{n-4}} \leq \frac{1}{\sqrt[n-4]{n^n}}\sqrt[n-4]{\frac{4^4a^{n-4}}{(a+b)^4(a+c)^4}}.

We must see now that

\displaystyle \sqrt[n-4]{\frac{4^4a^{n-4}}{(a+b)^4(a+c)^4}}=\sqrt[n-4]{\left(\frac{2a}{a+b}\right)^4\cdot \left(\frac{2a}{a+c}\right)^4\cdot a^{n-12}}.

From the above result we can apply once again the AM – GM Inequality and acquire

\displaystyle \sqrt[n-4]{\left(\frac{2a}{a+b}\right)^4\cdot \left(\frac{2a}{a+c}\right)^4\cdot a^{n-12}}\leq \frac{4\frac{2a}{a+b}+4\frac{2a}{a+c}+(n-12)a}{n-4}.

Summing up the \displaystyle 3 Inequalities together we get that

\displaystyle \sum_{cyc}\left(\frac{1}{3a+bc+\frac{n-4}{a}}\right)^{\frac{n}{n-4}}\leq \frac{1}{\sqrt[n-4]{n^n}}\cdot\frac{4\sum_{cyc}\left(\frac{2a}{a+b}+\frac{2a}{a+c}\right)+3(n-12)}{n-4}.

Moreover, we have that

\displaystyle \frac{1}{\sqrt[n-4]{n^n}}\cdot\frac{4\sum_{cyc}\left(\frac{2a}{a+b}+\frac{2a}{a+c}\right)+3(n-12)}{n-4}=\frac{1}{\sqrt[n-4]{n^n}}\cdot\frac{3\cdot 4\cdot 2+3(n-12)}{n-4},

which reduces to the \displaystyle \frac{3}{\sqrt[n-4]{n^n}}, Q.E.D.

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